Class 11 Engineering Maths - Principle Of Mathematical Induction

If $A = \left( {\begin{array}{{20}{c}}{\cos \theta }&{i\sin \theta }\\{i\sin \theta }&{\cos \theta }\end{array}} \right)$ , where $i = \sqrt { - 1}$ , then by the principle of Mathematics Induction prove that ${A^n} = \left( {\begin{array}{{20}{c}}{{\mathop{\rm cosn}\nolimits} \theta }&{i{\mathop{\rm sinn}\nolimits} \theta }\\{i{\mathop{\rm sinn}\nolimits} \theta }&{{\mathop{\rm cosn}\nolimits} \theta }\end{array}} \right)$

$\begin{array}{l}A = \left( {\begin{array}{*{20}{c}}{\cos \theta }&{i\sin \theta }\\{i\sin \theta }&{\cos \theta }\end{array}} \right)\\i = \sqrt { - 1} \\{A^n} = \left( {\begin{array}{*{20}{c}}{{\mathop{\rm cosn}\nolimits} \theta }&{i{\mathop{\rm sinn}\nolimits} \theta }\\{i{\mathop{\rm sinn}\nolimits} \theta }&{{\mathop{\rm cosn}\nolimits} \theta }\end{array}} \right)\\{A^n} = {{\mathop{\rm cosn}\nolimits} ^2}\theta - {i^2}{\sin ^2}n\theta = 1\end{array}$
Or else solve by putting

$\begin{array}{l}n = 1,2,3,4........\\{A^2} = \left( {\begin{array}{*{20}{c}}{\cos 2\theta }&{i\sin 2\theta }\\{i\sin 2\theta }&{\cos 2\theta }\end{array}} \right)\\{A^2} = \cos {2^2}\theta - {i^2}{\sin ^2}2\theta = 1\end{array}$

Hence Prooved

Using the principle mathematical induction , prove that $\forall \, n \, \epsilon \, N$
$I_n \, = \, \int_{0}^{\pi/2}cos^n \, \, x sin\, \, nx \,dx$
$= \, \dfrac{1}{2^{n \, + \, 1}}\left [2 \, + \, \dfrac{2^2}{2}\, +\dfrac{2^3}{3} .... \, + \, \, \dfrac{2^n}{n} \right ]$

p (1)

$\int_{0}^{\pi/2}$ cos x sin x dx

$= \, \dfrac{1}{2}[sin^2 \, x]_0^{\pi/2} \, = \, \dfrac{1}{2} \, = \, \dfrac{1}{2^{1 \, + \, 1}} \left[\dfrac{2}{1} \right]$
Assume p (m) to hold good.
p(m + 1) =

$\int_{0}^{\pi/2}$

$cos^{m+1}$ x sin (m + 1) x dx
=

$\int_{0}^{\pi/2}$

$cos^m$ x[sin (m + 1) x cos x] dx
=

$\int_{0}^{\pi/2}$

$cos^m$ x[cos(m + 1) x sin x + sin mx]
Using the formula of sin (A - B)
i.e., sin mx = sin [(m + 1)x - x]
= sin (m + 1) x cos x - cos (m + 1) x sin x....

$therfore$ p (m + 1) =

$\int_{0}^{\pi/2}$

$(cos^m$ x sin x )cos (m + 1) x + p (m), by
Integrating by parts.

$= \, \left [ \dfrac{cos^{m \, + \, 1}x}{m \, + \, 1} \, cos \, (m \, + \, 1)x\right ] \, + \, \dfrac{1}{m \, + \, 1}\int cos^{m \, + \, 1} \, x \, \times$ -(m + 1) sin (m + 1) x + p (m)
p(m + 1) =

$\dfrac{1}{m \, + \, 1}$ - p(m + 1) + p(m)
2p (m + 1) =

$\dfrac{1}{m \, + \, 1} \, + , \dfrac{1}{2^{m \, + \, 1}} \left[2 +\, + \, \dfrac{2^2}{2} \, + \, ... \, + \, \dfrac{2^m}{m} \right]$

$= \, \dfrac{1}{2^{m \, + \, 1}} \left[2 \, + \, \dfrac{2^2}{2} \, + \, ... \, + \, \dfrac{2^m}{m} \, + \, \dfrac{2^{m \, + \, 1}}{m \, + \, 1} \right]$

$\therefore$ p(m + 1) =

$\dfrac{1}{2^{m \, + \, 2}} \left[2 \, + \, \dfrac{2^2}{2} \, + \, ... \, + \, \dfrac{2^{m \, + \, 1}}{m \, + \, } \right]$

Prove that $2^n>n$ for all positive integers $n$

Let

$P(n):2^n>n$
When

$n=1,2^1>1$ .Hence

$P(1)$ is true.
Assume that

$P(k)$ is true for any positive integer

$k$ ,i.e.,

$2^k>k$
we shall now prove that

$P(k+1)$ is true whenever

$P(k)$ is true.
Multiplying both sides of

$(1)$ by

$2$ , we get

$2.2^k>2k$
i.e.,

$2^{k+1}>2k$

$k+k>k+1$

$\therefore 2^{k+1}>k+1$

For any natural number n , prove that inequality
|sin n x| $\geq$ n | sin x |

The inequality clearly holds for n = 1
We now assume
|sin mx|

$\leq$ | sin mx | + | sin x |
[

$\therefore$ | cos x |

$\leq$ 1 and | cos mx |

$\leq$ 1]

$\leq$ (m + 1) | | sin x |.
Hence by induction, the required inequality holds for every positive integer n

Prove that 7 is a factor of $2^{3n}$ - 1 for all natural numbers n.

Let P(n): 7 is a factor of

$2^{3n}$ - 1 be the given statement
Step 1: When n = 1

$2^{3(1)}$ - 1 = 7 and 7 is a factor of itself

$\therefore$ P(n) be true for n = k.
Step 2: Let P(n) be true for n = k.

$\Rightarrow$ 7 is a factor of

$2^{3k}$ - 1.

$\Rightarrow$

$2^{3k}$ - 1 = 7M, where M

$\epsilon$ N.

$\Rightarrow$

$2^{3k}$ = 7M + 1

$\rightarrow$ (1)
Now consider

$2^{3(k + 1)}$ - 1 =

$2^{3k + 3}$ - 1 =

$2^{3k}$ .

$2^{3}$ - 1
= 8(7M + 1) 1 (using (1)) = 56M + 7 (As

$2^{3k}$
= (7M + 1)

$\therefore$

$2^{3(k + 1)}$ - 1 = 7 (8m + 1)

$\Rightarrow$ 7 is factor of

$2^{3(k + 1)}$ - 1

$\Rightarrow$ P(n) istrue for n = k + 1

$\therefore$ By the principle of mathematical induction,
P(n) is true for all natural numbers n.
Hence, 7 is a factor

$2^{3n}$ - 1 for all n

$\epsilon$ N.

Prove the assertions of the following problems
Prove that $2^n > n^2$ for any natural number $n > 5.$

$n>5$

$\implies (n-1)^{2}>2$

$\implies n^{2}-2{n}-1>0$

$\implies 2 n^{2}-2{n}-1>n^{2}$

$\implies 2{n^{2}}>n^{2}+2{n}+1$

$\implies 2{n^{2}}>(n+1)^{2}$

Let us assume

$2^{n}>n^{2}$

$2^{n+1}=2.2^{n}>2{n^{2}}>(n+1)^{2}$ which is true

By mathematical induction

$2^{n}>n^{2},when\ (n)>5$

If $P(n)$ is the statement $''2^n\ge 3n"$ and if $P(r)$ is true prove that $P(r+1)$ is true.

1205469_1036695_ans_6e9333cddb5641f7a86371275cd2cdee.JPG

Using the principle of mathematical induction prove that $2 + 4 + 6 + .... + 2n = {n^2} + n$ .

$\begin{array}{l}2 + 4 + 6 + .... + 2n = {n^2} + n\\let\,\,\,n = 1\\\therefore 2 \times 1 = {1^2} + 1\\2 = 2\left( {true} \right)\\Let\,given\,equation\,be\,true\,for\,n = k\\\therefore 2 + 4 + 6 + .... + 2k = {k^2} + k\,\,\,\,\,\,\left( 1 \right)\\Induction\,step\\let\,n = k + 1\\\therefore \,2 + 4 + 6 + ... + 2\left( {k + 1} \right) = {\left( {k + 1} \right)^2} + \left( {k + 1} \right)\\2 + 4 + 6 + ..... + 2k + 2 = {k^2} + 1 + 2k + k + 1\\ = {k^2} + k + 2k + 2\\2\left( {1 + 2 + 3 + ... + \left( {k + 1} \right)} \right) = {k^2} + 3k + 2\\2\left[ {\dfrac{{k + 1}}{2}\left( {k + 1 + 1} \right)} \right]\\ \Rightarrow \left( {k + 1} \right)\left( {k + 2} \right) = {k^2} + 3k + 2\\ \Rightarrow {k^2} + 3k + 2 = {k^2} + 3k + 2\\Hence\,\,\,proved\end{array}$

Prove that $\left( {1 \times 2 \times 5} \right) + \left( {2 \times 3 \times 7} \right) + \left( {3 \times 4 \times 9} \right) + .....$ is $\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 7} \right)}}{6}$

$(1\times2\times5)+(2\times3\times7)+(3\times4\times9)+\cdots$

$p(n)=(1\times2\times5)+(2\times3\times7)+\cdot +[n\times(n+1)\times (2n+3)]$

$n=1$

${\rm Lhs}=1\times (2)\times (5)=10$

${\rm Rhs}=\dfrac{1(1+1)(1+2)(3+7)}{6}$

$=\dfrac{2\times3\times10}{6}=10$

$\hbox{true for }n=1$

$\hbox{then true for }k,$

${\rm so}\, p(k)=(1\times2\times5)+\cdot +k(k+1)(2k+3)$

$=\dfrac{k(k+1)(k+2)(3k+7)}{6}$

${\rm Now}\,p(k+1)=(3k^4+16k^3+27k^2+14k)/6$

$=[(1\times2\times5)+(2\times3\times7)+\cdots +k(k+1)(2k+3)]$

$+(k+1)(k+2)(2k+5)$

$=\dfrac{k(k+1)(k+2)(3k+7)}{6}+(k+1)(k+2)(2k+5)$

$=\dfrac{3k^4+16k^3+27k^2+14k+12k^3+66k^2+114k+60}{6}$

$=\dfrac{3k^4+28k^3+93k^2+128k+60}{6}$

$\hbox{hence true for }k+1$

$\therefore p(n)\hbox{ is true }\forall n\in IN$

Prove by mathematical induction,
${ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+....+{ n }^{ 2 }=\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 }$

$P(n):$

$1^2+2^2+3^2+........+n^2=\dfrac{n(n+1)(2n+1)}{6}$

$P(1):$

$1^2=\dfrac{1(1+1)(2(1)+1)}{6}$

$1=\dfrac{6}{6}=1$

$\therefore LHS=RHS$

Assume P(k) is true

$P(k):$

$1^2+2^2+3^2+........+k^2=\dfrac{k(k+1)(2k+1)}{6}$

P(k+1) is given by,

$P(k+1):$

$1^2+2^2+3^2+........+(k+1)^2=\dfrac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$

$\Rightarrow (k+1)\dfrac{k(2k+1)+6(k+1)}{6}=\dfrac{(k+1)(k+2)(2k+3)}{6}$

$\Rightarrow (k+1)\dfrac{2k^2+7k+6}{6}=\dfrac{(k+1)(k+2)(2k+3)}{6}$

$\Rightarrow \dfrac{(k+1)(k+2)(2k+3)}{6}=\dfrac{(k+1)(k+2)(2k+3)}{6}$

True for P(k+1)

Hence by Principle of mathematical induction

$1^2+2^2+3^2+........+n^2=\dfrac{n(n+1)(2n+1)}{6}$ is true for $\forall\ n\in N$

Using the principle of mathematical induction, show that;
$2+4+6+.....+2n=n^2+n$

Given; P(n) is 2 + 4 + 6 + …+ 2n = n² + n.

P(0) = 0 = 0²+ 0 ; it’s true.

P(1) = 2 = 1² + 1 ; it’s true.

P(2) = 2 + 4 = 2² + 2 ; it’s true.

P(3) = 2 + 4 + 6 = 3² + 2 ; it’s true.

Let P(k) be 2 + 4 + 6 + …+ 2k = k² + k is true;

⇒ P(k+1) is 2 + 4 + 6 + …+ 2k + 2(k+1) = k² + k + 2k +2

= (k² + 2k +1) + (k+1)

= $(k + 1)^2 + (k + 1)$
⇒ P(k+1) is true when P(k) is true.

∴ By Mathematical Induction $2 + 4 + 6 + …+ 2n = n^2 + n$ is true for all natural numbers n.

Using Principle of mathematical induction prove that $6^n -1$ divisible by $5$ .

$\begin{array}{l}P(n) = {6^n} - 1\\P(1) = {6^1} - 1 = 5\,is\,divisible\,by\,5\\Let,P(k)\,is\,divisible\,by\,5\\{6^n} - 1 = 5m......(1)\\P(k + 1) = {6^{k + 1}} - 1 = {6.6^k} - 1\\ = (5 + 1){.6^k} - 1 = 5 \times {6^k} + {6^k} - 1\\ = 5 \times {6^k} + 5m = 5({6^k} + m)\\P(k + 1)\,is\,divisible\,by\,5\\So,{6^n} - 1\,is\,always\,divisible\,by\,5.\end{array}$

Using the principle of Mathematical Induction, prove the following for all $n$ and $N$
1) ${3}^{n}>{2}^{n}$

Let

$P\left( n \right):{3^n} > {2^n}$

Step (i) for

$n=1$

$P(1) : 3>2$

which is true for

$n=1$

Step (ii)

Let it is true for

$n=k$

So,

${3^k} > {2^k}$

Step (iii)

We have to prove that it is true for

$n=k+1$ using step (ii)

Now

$3^{k+1}=3^k \times 3$

$3^{k+1}>2^k \times 3$ using step (ii)

$3^{k+1}>2^k (2+1)$

$3^{k+1}>2^{k+1}+2^k$

Therefore,

$3^{k+1}>2^{k+1}$

So it is true for

$n=k+1$

therefore, it is true for all

$n \in N$

Hence proved.

Prove that the square of any positive integer of the form 5m+1 will leave a remainder 1 when divided by 5 for some integer m

Given a positive integer of the form

$5m+1$

Say

$N=5m+1\\ { N }^{ 2 }=25{ m }^{ 2 }+1+10m\\ =25{ m }^{ 2 }+10m+1\\ =5(5{ m }^{ 2 }+2m)+1\\ =5k+1$

where

$k=5{ m }^{ 2 }+2m=$ same integer

${ N }^{ 2 }=5k+1$

From the above we can infer that

${ N }^{ 2 }$ , of the form

$5k+1$ , leaves a remainder of

$'1'$ when divided by

$5$ .

Prove using PMI
$\left( {1 + \frac{3}{1}} \right)\left( {1 + \frac{5}{4}} \right)\left( {1 + \frac{7}{9}} \right)...\left( {1 + \frac{{\left( {2n + 1} \right)}}{{{n^2}}}} \right) = {\left( {n + 1} \right)^2}$

$\left( {1 + \cfrac{3}{1}} \right)\left( {1 + \cfrac{5}{4}} \right)\left( {1 + \cfrac{7}{9}} \right)...\left( {1 + \cfrac{{\left( {2n + 1} \right)}}{{{n^2}}}} \right) = {\left( {n + 1} \right)^2}$

for

$n = 1,1 + \cfrac{3}{1} = \cfrac{4}{1} = {\left( {1 + 1} \right)^2}$

Let the statement be true for

$n=k$

$\Rightarrow \left( {1 + \cfrac{3}{1}} \right)\left( {1 + \cfrac{5}{4}} \right)\left( {1 + \cfrac{7}{9}} \right)...\left( {1 + \cfrac{{\left( {2k + 1} \right)}}{{{k^2}}}} \right) = {\left( {k + 1} \right)^2}$

for

$n=k+1$ ,

$\begin{array}{l}\left( {1 + \cfrac{3}{1}} \right)\left( {1 + \cfrac{5}{4}} \right)\left( {1 + \cfrac{7}{9}} \right)...\left( {1 + \cfrac{{\left( {2k + 1} \right)}}{{{k^2}}}} \right)\left( {1 + \cfrac{{\left( {2k + 3} \right)}}{{{{\left( {k + 1} \right)}^2}}}} \right)\\ = {\left( {k + 1} \right)^2}\left( {1 + \cfrac{{\left( {2k + 3} \right)}}{{{{\left( {k + 1} \right)}^2}}}} \right)\\ = {\left( {k + 1} \right)^2}\left( {\cfrac{{{k^2} + 2k + 1 + 2k + 3}}{{{{\left( {k + 1} \right)}^2}}}} \right)\\ = {\left( {k + 2} \right)^2}\end{array}$

$\therefore$ The statement is true by PMI

If $P(n)$ is the statement $n^{2}-n+41$ is prime, prove that $P(1),P(2)$ and $P(3)$ are true.
Prove also that $P(41)$ is not true.

$x^2-x+41$

$P(n)=x^2-x+41$ is a prime number

$P(1)=1^2-1+41$ is a prime No. of

$P(1)=41$ is prime, which is true

$P(2)= 2^2-2+41$ is a prime No, i.e.,

$P(2)=43$ is prime No, which is true

$P(211)=41^2-41+41$ is a prime No,

i.e.,

$P(41)=41^2$ is a primes which is Not true.

Solve: $(2n+7)<(n+3)^2$

$2n+7<{ \left( n+3 \right) }^{ 2 }$

$\Rightarrow 2n+7<{ n }^{ 2 }+9+6n$

$\Rightarrow { n }^{ 2 }+4n+2>0$

Now, adding and subtracting

$4$ we get,

${ n }^{ 2 }+4n+2+4-4>0$

$\Rightarrow { n }^{ 2 }+4n+4-4+2>0$

$\Rightarrow { (n }^{ 2 }+2\times n\times 2+{ 2 }^{ 2 })-2>0$

$\Rightarrow { \left( n+2 \right) }^{ 2 }>2$

$\Rightarrow n+2>\sqrt { 2 }$

$\Rightarrow n>\sqrt { 2 } -2$

Using P.M.I. prove that
${n^2} < {2^n}\forall n \ge 5$

Given, ${n}^{2}< {2}^{n} , \forall n\ge 5$

By PMI,

1) Let $n=5$

=> ${5}^{2} < {2}^{5}$

=> $25<32$ , which is TRUE mathematically.

2) Let n=k

=> ${k}^{2} < {2}^{k}$ , Assumed TRUE $\forall k\ge 5$

3) Let $n=k+1$

=> ${(k+1)}^{2} < {2}^{k+1} k>=5$

=> ${k}^{2}+2k+1 < {2}^{k+1}$

=> $\dfrac { { k }^{ 2 } }{ 2 } + k + \dfrac { 1 }{ 2 } < {2}^{k}$

Now, we have already assumed that ${k}^{2} < {2}^{k}$ ,

which obviously means that $\dfrac { { k }^{ 2 } }{ 2 } < {2}^{k}$

Therefore, we just have to show that:

$k +\frac { 1 }{ 2 } < \dfrac { { k }^{ 2 } }{ 2 }$ (the other half of ${k}^{2}$ ), $\forall k\ge 5$ .

Taking all the terms to RHS, we get

$\dfrac { { k }^{ 2 } }{ 2 } – k – \dfrac { 1 }{ 2 } > 0$

For checking that the above inequation is true $\forall k\ge 5$ , we begin finding the roots of the quadratic equation.

They come out to be $1\pm \sqrt { 2 }$

Since $k=5$ is greater than both the roots (where the equation is zero-valued), and the coefficient of ${ k }^{ 2 }$ is positive (which tells us if the quadratic function is increasing or decreasing on either side of the roots on the number line, which in case of positive coefficient is increasing and is decreasing for negative coefficient) we can safely say that the above inequation holds TRUE.

Hence, we have proved by PMI that ${n}^{2}< {2}^{n}, for all n\ge 5$

$x = \sin t\,\,\,\,\,\,\,\,\,\,y = \cos mt$
Prove $\left( {1 - {x^2}} \right){y_{n + 2}} - \left( {2n - 1} \right){y_{n + 1}} - \left( {{n^2} - {m^2}} \right){y_n} = 0$

$x=\sin t y=\cos mt$

$t=\sin^{-1}(x)$

$y=\cos (m\sin^{-1}(x))$

$(1-x^{2})y_{n+2}-(2n+1)xy_{n+1}+(m^{2}-n^{2})y_{n}=0$

$\dfrac{dy}{dx}=-\sin(m\sin^{-1}(x))=\dfrac{m}{\sqrt{1-x^{2}}}$

$\sqrt{1-x^{2}}\dfrac{dy}{dx}=-m\sin (m\sin^{-1}x)$

$(1-x^{2})\left(\dfrac{dy}{dx}\right)^{2}=-m^{2}\left[\sin(m\sin^{-1}x))\right]^{2}$

$=m^{2}\left[1-\cos^{2}(m\sin^{-1}x)\right]$

$(1-x^{2})\left(\dfrac{dy}{dx}\right)^{2}=m^{2}[1-y^{2}]$

$\left| \begin{matrix} \dfrac { dy }{ dx } ={ y }_{ 1 } \\ \dfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ y }_{ 2 } \end{matrix} \right.$

$(1-x^{2})2y_{1}y_{2}+y_{1}^{2}(-2x)=m^{2}(-2y)y_{1}$

$(1-x^{2})y_{2}-xy_{1}=-m^{2}y$

$\therefore (1-x^{2})y_{2}-xy_{1}+m^{2}y=0$

$(1-x^{2})y_{m+2}+m(-2x)_{y m+1}\neq \dfrac{m(n-1)}{2}. (-2)y_{m}$

$xy_{x+1}-xy_{m}-m^{2}y_{n}=0$

$(1-x^{2})y_{n+2}-x(2x++1)Y_{n+1}+(n^{2}+n-n+m^{2})yx=0$

$(1-x^{2})y_{m+2}-(2n+1)xy_{x+1}+(m^{2}-n^{2})yx=0$

Use the principle of mathematical induction to show that :
$2 + 2^2 + ..... + 2^n = 2^{n+1} - 2$ for every natural number $n.$

$\begin{array}{l} p\left( n \right) :2+{ 2^{ 2 } }+..............+{ 2^{ n } }={ 2^{ n+1 } }-2 \\ When,\, \, n=1 \\ L.H.S=2\, \, and\, \, R.H.S=2 \\ so,\, \, for\, \, n=1\, \, p\left( n \right) \, \, is\, \, true. \\ Let\, \, P\left( R \right) \, \, :2+{ 2^{ 2 } }+{ ....2^{ k } }={ 2^{ k+1 } }-2 \\ Therefore, \\ p\left( { k+1 } \right) :2+{ 2^{ 2 } }+{ ....2^{ k+1 } }={ 2^{ k+2 } }-2 \\ ={ 2^{ k+1 } }.2-2 \\ =2\left( { { 2^{ k+1 } }-1 } \right) \\ So,\, \, p\left( { k+1 } \right) \, \, is\, \, also\, \, true. \\ Hence\, \, by\, \, the\, \, principle\, \, of\, \, mathematical\, \, induction, \\ p\left( n \right) \, \, is\, \, true\, \, for\, \, all\, \, n\in N. \end{array}$

Show that only one of the no $n, n + 2$ and $n + 4$ in divisible $3$ .

We applied Euclid Division algorithm on

$n$ and

$3$ .

$a = bq +r$ on putting

$a = n$ and

$b = 3$

$n = 3q +r$ ,

$0<r<3$
i.e

$n = 3q$ -------- (1),

$n = 3q +1$ --------- (2),

$n = 3q +2$ -----------(3)

$n = 3q$ is divisible by

$3$
or

$n +2 = 3q +1+2 = 3q +3$ also divisible by

$3$
or

$n +4 = 3q + 2 +4 = 3q + 6$ is also divisible by

$3$
Hence

$n, n+2 , n+4$ are divisible by

$3$ .

Prove by method of induction, for all $n \in$
$1^2 + 2^2 + 3^2 + ... + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}$ .

Let

$P\left( n \right) = {1^2} + {2^2} + {3^2} + ..... + {n^2} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$

For

$n=1$ ,

$L.H.S={1^2}=1$

$R.H.S=\frac{{1\left( {1 + 1} \right)\left( {2 \times 1 + 1} \right)}}{6} = \frac{{1 \times 2 \times 3}}{6} = 1$

Hence,

$L.H.S=R.H.S$

$\therefore P(n)$ is true for

$n=1$ .

If ${a_1},\,{a_2},\,{a_3},\,...,{a_n}$ are in A.P. and ${a_i} > 0$ for all i, then show that $\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_3}}} + ... + \dfrac{1}{{{a_{n - 1}}{a_n}}} = \dfrac{{n - 1}}{{{a_1}{a_n}}}$ .

Since for all i,

$a_i>0$ , therefore above result can be proved by using principle of mathematical induction.

Step 1: Checking for n=2,

$\dfrac{1}{a_{1}a_{2}}=\dfrac{1}{a_{1}a_{2}}$ which is true.

Step 2: Assuming the given relation to be true for

$1<=m<n$ ,

$\dfrac{1}{a_{1}a_{2}}+\dfrac{1}{a_{2}a_{3}}+$ so on upto

$+\dfrac{1}{a_{m-1}a_{m}}=\dfrac{m-1}{a_{1}a_{m}}$

Step 3: To prove that result is true for next natural number m+1.

let d be the common difference

we have

$\dfrac{1}{a_{1}a_{2}}+\dfrac{1}{a_{2}a_{3}}+$ so on upto

$\dfrac{1}{a_{m-1}a_{m}}+\dfrac{1}{a_{m}a_{m+1}}=\dfrac{m-1}{a_{1}a_{m}}+\dfrac{1}{a_{m}a_{m+1}}$

$\rightarrow\dfrac{1}{a_{m}}*\dfrac{ma_{m+1}-a_{m+1}-a_{1}}{a_{1}a_{m+1}}$

$\rightarrow\dfrac{1}{a_{m}}\dfrac{ma_{m+1}-(a_{1}+md)-a_{1}}{a_{1}a_{m+1}}$

on further simplification, we get

$\dfrac{m}{a_{1}a_{m+1}}$ which is true for m+1 and hence for all n.

Hence

$\dfrac{1}{a_{1}a_{2}}+\dfrac{1}{a_{2}a_{3}}+$ so on upto

$+\dfrac{1}{a_{n-1}a_{n}}=\dfrac{n-1}{a_{1}a_{n}}$

To prove: $n \le 2^{n}$ for $n\ in\ N$
$P(n):n \le 2^{n}.........$ for $n\ in\ N$

1171108_1313274_ans_b471b31428194a7982ff2affbe7fc6cc.jpg

Can the number $6^{n}$ , where n is a natural number, end with digit 5? Give reasons.

If a number ends with 5, then it is not divisible by 2. However, 6 is divisible by 2, and so every power of 6. That's why no such power can end with 5.

For every natural number $n(>1)$ , prove that $4^{n}+15n-1$ is divisible by $9$ .

Proving by induction,

for

$n=1$ ,

$4^n+15n-1=18$

Let for

$n=k$ ,

$4^k+15k-1=9\lambda_1$

for

$n=k H=4^{k+1}+15(k+1)-1$

$=4.4^k+15k+14$

$=4(9\lambda_1+1-15k)+15k+14$

$=36\lambda_1+4-60k+15k+14=36\lambda_1-45k+18$

$=9(4\lambda_1-5k+2)=9\lambda_1$ .

1208040_1395146_ans_dc46f5a8e86c4b6ca358cd70bd2e34f0.jpg

Prove by method of induction; for all $n \in N$ $3 + 7 + 11 + \ldots . . . +$ to $n$ terms $= n ( 2 n + 1 )$

$3+7+11+15+...……..n$ terms

$=n(2n+1)$

Put

$n=1$

LHS

$=3$

RHS

$=1(2(1)+1)=2+1=3$

LHS

$=$ RHS

Let us assume that the result is true for

$n=k$

$3+7+11+15+...….+(4k-1)=k(2k+1)$

Let us prove this result for

$n=k+1$

$3+7+11+15+...……+(4k-1)+(4k+3)=(k+1)(2k+3)$

LHS

$=3+7+11+...………+(4k-1)+(4k+3)$

$=k(2k+1)+(4k+3)$ (By assumption for k)

$=2k^2+k+4k+3$

$=2k^2+3k+2k+3$

$=k(2k+3)+1(2k+3)=(k+1)(2k+3)$

$=$ RHS.

So, the result is true for

$n=k+1$

Hence by induction principle, the given result is true for all natural numbers.

1215918_1328102_ans_8c90d3cd53bf44baa6fdb6da90f3ded9.jpg

Prove that $3^{n+1} > 3(n+1)$ .

Let

$P(n) : 3^{n+1}> 3(n+1)$

Step 1 :

Put

$n=1$

Then,

$3^{2} > 3(2)$

$\implies P(n)$ is true for

$n=k$

Step 2 :

Assume that P(n) is true for

$n=k$

Then,

$3^{k+1} > 3(k+1)$

Multiplying throughout by 3

$3^{k+1}.3>3(k+1).3=9k+9=3(k+2)$

$3^{\overline {k+1} +1} > 3(\overline {k+1} +1)$

$P(n)$ is true for

$n=k+1$ .

Therefore, by the principle of mathematical induction, P(n) is true for all

$n\in N$ .

Hence,

$3^{n+1} > 3(n+1)$ , for every

$n\in N$ .

"The product of three consecutive positive integers is divisible by 6". Is.this statement true or false"? Justify your answer.

It is

$True$
Let

$a= n\left(n+1\right)\left(n+2\right)$ ..1
Because out of three consecutive positive integers , one of the integers must be an even integer and one integer as multiple of 3.
let

$n=2p \ and \ n+1 = 3q$

$a = 2p\left(3q\right)\left(n+2\right) = 6 \left[\left(pq\right)\left(n+2\right)\right] = 6m \ where \ m = \left[\left(pq\right)\left(n+2\right)\right]$

Prove that $1 + 2 + 3 + ..... + n =$ $\displaystyle \frac {n(n\, +\, 1)}{2}$ .

Let P(n): 1 + 2 + ..... + n =

$\displaystyle \frac {n(n\, +\, 1)}{2}$ be the given statement.
Step 1: Put n = 1
Then, L.H.S. = 1 and R.H.S. =

$\displaystyle \frac {1(1\, +\, 1)}{2}\, =\, 1$

$\therefore$ L.H.S = R.H.S.
Step 2: Assume that P(n) is true for n = k.

$\therefore\, 1\, +\, 2\, +\, 3\, .....\, +\, k\, =\, \displaystyle \frac {k(k\, +\, 1)}{2}$
Adding (k + 1) on both sides we get

$1\, +\, 2\, +\, 3\, .....\, +\, k\,\, +\, (k\, +\, 1)\, =\, \displaystyle \frac {k(k\, +\, 1)}{2}$

$=\, \left ( \displaystyle \frac {k}{2}\, +\, 1 \right )$

$=\, \displaystyle \frac {(k\, +\, 1)(k\, +\, 2)}{2}$

$=\, \displaystyle \frac {(k\, +\, 1)(\overline{k\, +\, 1}\, +\, 1)}{2}$

$\Rightarrow\, P(n)$ is true n = k + 1

$\therefore$ By the principle of mathematical induction
p(n) istrue for all natural numbers n.
Hence, 1 + 2 + 3 + ..... + n =

$\displaystyle \frac {n(n + 1)}{2}$ for all n

$\epsilon$ N

Prove that $3^{n + 1} > 3(n + 1)$ .

Let P(n):

$3^{n + 1}$ > 3(n + 1)
Step 1: Put n = 1
Then, 3

$^2$ > 3(2)

$\Rightarrow$ p(n) is true for n = 1
Step 2: Assume that p(n) is true for n = k then,

$3^{k + 1}$ > 3(k + 1)
Multiplying throughout with '3'.

$3^{k + 1}\times 3> 3(k + 1) \times 3 = 9k + 9 = 3(k + 2) + (6k + 3) > 3(k + 2)$

$\Rightarrow\, 3^{\overline{k\, +\, 1}\, +\, 1}\, >\, 3(\overline{k\, +\, 1}\, +\, 1)$
P(n) is true for n = k + 1

$\therefore$ By the principle of mathematical induction,
P(n) is true for all n

$\epsilon$ N.
Hence,

$3^{n + 1}$ > 3(n + 1)

$\forall$ n

$\epsilon$ N.

Prove $1+3+5+\cdots+(2n-1)=n^2$

Let P(n):

$1+3+5+\cdots+(2n-1)=n^2$ be the given statement
Step 1: Put

$n=1$
Then, L.H.S

$=1$
R.H.S

$=1^2=1$

$\therefore$ L.H.S

$=$ R.H.S

$\implies$ P(n) is true for

$n=1$
Step 2: Assume_that P(n) is true for

$n=k$

$\therefore1+3+5+\cdots+(2k-1)=k^2$

Adding

$2k+1$ on both sides, we get

$1+3+5+\cdots+(2k-1)+(2k+1)=k^2+(2k+1)=(k+1)^2$

$\therefore1+3+5+\cdots+(2k-1)+(2(k+1)-1)=(k+1)^2$

$\implies$ P(n) is true for

$n=k+1$

$\therefore$ By the principle of mathematical induction P(n) is true for all natural numbers 'n'.
Hence

$1+3+5+\cdots+(2n-1)=n^2$ , for all

$n\in N$

Prove $3^{n+1}>$ $3(n+1)$

Let P(n):

$3^{n+1}>3(n+1)$
Step 1: Put

$n=1$
Then,

$3^2>3(2)$

$\implies$ p(n) is true for

$n=1$
Step 2: Assume that P(n) is true for

$n=k$
Then,

$3^{k+1}>3(k+1)$
Multiplying throughout with '3'

$3^{k+1}\times3>3(k+1)\times3=9k+9=3(k+2)+(6k+3)>3(k+2)$

$\implies3^{\displaystyle\overline{k+1}+1}>3(\overline{k+1}+1)$

P(n) is true for

$n=k+1$

$\therefore$ By the principle of mathematical induction, P(n) is true for all

$n\in N$
Hence,

$3^{n+1}>3(n+1)$ .

Prove the following using the principle of mathematical induction for all $n\in N$ :
$1+3+{3}^{2}+........+{3}^{n-1}=\cfrac{({3}^{n}-1)}{2}$

Let the given statement be

$P(n)$ i.e.,

$P(n)=1+3+{3}^{2}+......+{3}^{n-1}=\cfrac{({3}^{n}-1)}{2}$
For

$n=1$ , we have

$P(1)=\cfrac{({3}^{1}-1)}{2}=\cfrac{3-1}{2}=\cfrac{2}{2}=1$ , which is true.
Let

$P(k)$ be true for some positive integer

$k$ , i.e.,

$1+3+{3}^{2}+........+{3}^{k-1}=\cfrac{({3}^{k}-1)}{2}..........(i)$
We shall now prove that

$P(k+1)$ is true.
Now

$P(k+1)=1+3+{3}^{2}+........+{3}^{k-1}+{3}^{(k-1)+1}$

$=(1+3+{3}^{2}+.....+{3}^{k-1})+{3}^{k}$

$=\cfrac{({3}^{k}-1)}{2}+{3}^{k}$ [Using

$(i)$ ]

$=\cfrac{{({3}^{k}-1)}+{2.3}^{k}}{2}$

$=\cfrac{3.{3}^{k}-1}{2}$

$=\cfrac{{3}^{k+1}-1}{2}$
Thus

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$ .

Prove that $1 + 3 + 5 + ..... + (2n - 1) = n$ $^2$ .

Let P(n): 1 + 3 + 5 + ..... + (2n - 1) = n

$^2$ be the given statement
Step 1: Put n = 1
Then, L.H.S = 1
R.H.S = (1)

$^2$ = 1

$\therefore$ . L.H.S = R.H.S.

$\Rightarrow$ P(n) istrue for n = 1
Step 2: Assume that P(n) istrue for n = k.

$\therefore$ 1 + 3 + 5 + ..... + (2k - 1) = k

$^2$
Adding 2k + 1 on both sides, we get
1 + 3 + 5 ..... + (2k - 1) + (2k + 1) = k

$^2$ + (2k + 1) = (k + 1)

$^2$

$\therefore$ 1 + 3 + 5 + ..... + (2k -1) + (2(k + 1) - 1) = (k + 1)

$^2$

$\Rightarrow$ P(n) is true for n = k + 1.

$\therefore$ by the principle of mathematical induction P(n) is true for all natural numbers 'n'
Hence, 1 + 3 + 5 + ..... + (2n - 1) =n

$^2$ , for all n

$\epsilon$ n

Prove the following by using principle of mathematical induction for all $n\in N: 1.3+3.5+5.7+.......+(2n-1)(2n+1)=\cfrac{n(4{n}^{2}+6n-1)}{3}$

Let the given statement be

$P(n)$ , i.e.,

$P(n) :1.3+3.5+5.7+.......+(2n-1)(2n+1)=\cfrac{n(4{n}^{2}+6n-1)}{3}$

For

$n=1$ , we have

$P(1):1.3=3=\cfrac{1(4.{1}^{2}+6.1-1)}{3}=\cfrac{4+6+1}{3}=\cfrac{9}{3}=3$ , which is true
Let

$P(k)$ be true for some

$k\in N$ , i.e.,

$1.3+3.5+5.7+......+(2k-1)(2k+1)=\cfrac {k(4{k}^{2}+6k-1)}{3}.........(i)$
We shall now prove that

$P(k+1)$ is true. i.e.

$1.3+3.5+5.7+.......+(2k-1)(2k+1)+\{2(k+1)-1\}\{2(k+1)+1\}=\cfrac { \left( k+1 \right) \left\{ 4{ (k+1) }^{ 2 }+6(k+1)-1 \right\} }{ 3 }$

Consider LHS,

$(1.3+3.5+5.7+.......+(2k-1)(2k+1)+\{2(k+1)-1\}\{2(k+1)+1\}$

$=\cfrac { k(4{ k }^{ 2 }+6k-1) }{ 3 } +(2k+2-1)(2k+2+1)$ [Using

$(i)$ ]

$=\cfrac { k(4{ k }^{ 2 }+6k-1) }{ 3 } +(2k+1)(2k+3)$

$=\cfrac { k(4{ k }^{ 2 }+6k-1) }{ 3 } +(4{ k }^{ 2 }+8k+3)$

$=\cfrac { k(4{ k }^{ 2 }+6k-1)+3(4{ k }^{ 2 }+8k+3) }{ 3 }$

$=\cfrac { 4{ k }^{ 3 }+6{ k }^{ 2 }-k+12{ k }^{ 2 }+24k+9 }{ 3 }$

$=\cfrac { 4{ k }^{ 3 }+18{ k }^{ 2 }+23k+9 }{ 3 }$

$=\cfrac { 4{ k }^{ 3 }+14{ k }^{ 2 }+9k+4{ k }^{ 2 }+14k+9 }{ 3 }$

$=\cfrac { \left( k+1 \right) \left( 4{ k }^{ 2 }+14k+9 \right) }{ 3 }$

$=\cfrac { \left( k+1 \right) \left\{ 4{ k }^{ 2 }+8k+4+6k+6-1 \right\} }{ 3 }$

$=\cfrac { \left( k+1 \right) \left\{ 4({ k }^{ 2 }+2k+1)+6(k+1)-1 \right\} }{ 3 }$

$=\cfrac { \left( k+1 \right) \left\{ 4{ (k+1) }^{ 2 }+6(k+1)-1 \right\} }{ 3 }$
Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$

Prove the following by using the principle of mathematical induction for all $n\in N: 1.3+2.{3}^{2}+3.{3}^{3}+.....+n.{3}{^n}=\cfrac{(2n-1){3}^{n+1}+3}{4}$

Let the given statement be

$P(n)$ i.e.,

$1.3+2.{3}^{2}+3.{3}^{3}+........n.{3}^{n}=\cfrac{(2n-1){3}^{n+1}+3}{4}$

$P(n)$ :
For

$n=1$ , we have

$P(1):1.3=3=\cfrac { \left( 2.1-1 \right) { 3 }^{ 1+1 }+3 }{ 4 } =\cfrac { { 3 }^{ 2 }+3 }{ 4 } =\cfrac { 12 }{ 4 } =3........(i)$
We shall now prove that

$P(k+1)$ is true.
Consider

$1.3+2.{3}^{2}+3.{3}^{3}+.......k.{3}^{k}+(k+1){3}^{k+1}=(1.3+2.{3}^{2}+3.{3}^{3}+.....+k.{3}^{k}+(k+1){3}^{k+1}$

$=\cfrac { (2k-1){ 3 }^{ k+1}+3}{4}+(k+1){3}^{k+1}$ [Using

$(i)$ ]

$=\cfrac { (2k-1){ 3 }^{ k+1 }+3+4(k+1){ 3 }^{ k+1 } }{ 4 } \quad \quad =\cfrac { { 3 }^{ k+1 }\left\{ 2k-1+4(k+1) \right\} +3 }{ 4 }$

$=\cfrac { { 3 }^{ (k+1)+1 }\left\{ 2k+1 \right\} +3 }{ 4 }$

$=\cfrac { \left\{ 2(k+1)-1 \right\} { 3 }^{ (k+1)+1 }+3 }{ 4 }$
Thus

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction. statement

$P(n)$ is true for all natural numbers i.e.,

$n$ .

Prove the following by using the principle of mathematical induction for all $n\in N:1.2.3+2.3.4+......+n(n+1)(n+2)=\cfrac{n(n+1)(n+2)(n+3)}{4}$

Let the given statement be

$P(n)$ , i.e.,

$P(n):1.2.3+2.3.4+.....+n(n+1)(n+2)=\cfrac{n(n+1)(n+2)(n+3)}{4}$
For

$n=1$ , we have

$P(1):1.2.3=6=\cfrac{1(1+1)(1+2)(1+3)}{4}=\cfrac{1.2.3.4}{4}=6$ , which is true.
Let

$P(k)$ be true for some positive integer

$k$ i.e.,

$1.2.3+2.3.4+.....k(k+1)(k+2)=\cfrac{k(k+1)(k+2)(k+3)}{4}..........(i)$
We shall now prove that

$P(k+1)$ is true.
Consider

$1.2.3+2.3.4+......+k(k+1)(k+2)+(k+1)(k+2)(k+3)=$

${1,2,3+2,3,4+.....k(k+1)(k+2)}+(k+1)(k+2)(k+3)$

$=\cfrac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)$ [Using

$(i)$ ]

$=(k+1)(k+2)(k+3)(\cfrac{k}{4}+1)$

$=\cfrac{(k+1)(k+2)(k+3)(k+4)}{4}$

$=\cfrac{(k+1)(k+1+1)(k+1+2)(k+1+3)}{4}$
Thus,

$P(k+1)$ is true whenever

$P(k)$ is true
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$

Prove the following by using the principle of mathematical induction for all $n \in N: \cfrac{1}{2}+\cfrac{1}{4}+\cfrac{1}{8}+.....+\cfrac{1}{{2}^{n}}=1-\cfrac{1}{{2}^{n}}$

Let the given statement be

$P(n)$ , i.e.,

$P(n):\cfrac{1}{2}+\cfrac{1}{4}+\cfrac{1}{8}+.....+\cfrac{1}{{2}^{n}}=1-\cfrac{1}{{2}^{n}}$
For

$n=1$ , we have

$P(1)=\cfrac{1}{2}=1-\cfrac{1}{{2}^{1}}=\cfrac{1}{2}$ , which is true.
Let

$P(k)$ be true for some

$k\in N$ , i.e.,

$\cfrac{1}{2}+\cfrac{1}{4}+\cfrac{1}{8}+.....+\cfrac{1}{{2}^{k}}=1-\cfrac{1}{{2}^{k}}......(i)$
We shall now prove that

$P(k+1)$ is true.
Consider

$\left( \cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 8 } +.....+\cfrac { 1 }{ { 2 }^{ k } } \right) +\cfrac { 1 }{ { 2 }^{ k+1 } }$

$=\left( 1-\cfrac { 1 }{ { 2 }^{ k } } \right) +\cfrac { 1 }{ { 2 }^{ k+1 } }$

$=1-\cfrac { 1 }{ 2^k } +\cfrac { 1 }{ { 2.2 }^{ k } }$

$=1-\cfrac { 1 }{ { 2 }^{ k } } \left( 1-\cfrac { 1 }{ 2 } \right)$

$=1-\cfrac { 1 }{ { 2 }^{ k } } \left( \cfrac { 1 }{ 2 } \right)$

$=1-\cfrac { 1 }{ { 2 }^{ k+1 } }$
Thus

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$

Prove the following by using the principle of mathematical induction for all $n\in N:\cfrac { 1 }{ 1.2.3 } +\cfrac { 1 }{ 2.3.4 } +\cfrac { 1 }{ 3.4.5 } +.....+\cfrac { 1 }{ n(n+1)(n+2) } =\cfrac { n(n+3) }{ 4(n+1)(n+2) }$

Let the given statement be

$P(n)$ i.e.,

$P(n):\cfrac { 1 }{ 1.2.3 } +\cfrac { 1 }{ 2.3.4 } +\cfrac { 1 }{ 3.4.5 } +.....+\cfrac { 1 }{ n(n+1)(n+2) } =\cfrac { n(n+3) }{ 4(n+1)(n+2) }$

For

$n=1$ , we have

$P(1):\cfrac{1}{1.2.3}=\cfrac{1.(1+3)}{4(1+1)(1+2)}=\cfrac{1.4}{4.2.3}=\cfrac{1}{1.2.3}$ , which is true.
Let

$P(k)$ be true for some

$k\in N$ i.e.,

$\cfrac { 1 }{ 1.2.3 } +\cfrac { 1 }{ 2.3.4 } +\cfrac { 1 }{ 3.4.5 } +.....+\cfrac { 1 }{ k(k+1)(k+2) } =\cfrac { k(k+3) }{ 4(k+1)(k+2) } ........(i)$
We shall now prove that

$P(k+1)$ is true.

$\cfrac { 1 }{ 1.2.3 } +\cfrac { 1 }{ 2.3.4 } +\cfrac { 1 }{ 3.4.5 } +.....+\cfrac { 1 }{ k(k+1)(k+2) } +\cfrac { 1 }{ (k+1)(k+2)(k+3) } =\dfrac { (k+1)(k+4) }{ 4(k+2)(k+3) }$

Consider

$\left[ \cfrac { 1 }{ 1.2.3 } +\cfrac { 1 }{ 2.3.4 } +\cfrac { 1 }{ 3.4.5 } +.....+\cfrac { 1 }{ k(k+1)(k+2) } \right] +\cfrac { 1 }{ (k+1)(k+2)(k+3)\quad \quad }$

$\quad =\cfrac { 1 }{ (k+1)(k+2) } \left\{ \cfrac { k(k+3) }{ 4 } +\cfrac { 1 }{ k+3 } \right\}$ [Using

$(i)$ ]

$=\cfrac { 1 }{ (k+1)(k+2) } \left\{ \cfrac { k{ \left( k+3 \right) }^{ 2 }+4 }{ 4(k+3) } \right\}$

$=\cfrac { 1 }{ (k+1)(k+2) } \left\{ \cfrac { k({ k }^{ 2 }+6k+9)+4 }{ 4(k+3) } \right\}$

$=\cfrac { 1 }{ (k+1)(k+2) } \left\{ \cfrac { { k }^{ 3 }+6{ k }^{ 2 }+9{ k }+4 }{ 4(k+3) } \right\}$

$=\cfrac { 1 }{ (k+1)(k+2) } \left\{\cfrac { k^3+k^2+5k^2+5k+4k+4 }{ 4(k+3) } \right\}$

$=\cfrac { 1 }{ (k+1)(k+2) } \left\{ \cfrac { (k+1)(k^2+5k+4) }{ 4(k+3) } \right\}$

$=\cfrac { 1 }{ (k+1)(k+2) } \left\{ \cfrac { (k+1)^2(k+4) }{ 4(k+3) } \right\}$

$=\cfrac { (k+4)(k+1) }{ 4(k+1)(k+2)(k+3) }$

$=\cfrac { (k+1)\left\{ (k+1)+3 \right\} }{ 4\left\{ (k+1)+1 \right\} \left\{ (k+1)+2 \right\} }$
Thus

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$

Prove the following by using the principle of mathematical induction for all $n\in$ :
$1+\cfrac{1}{(1+2)}+\cfrac{1}{(1+2+3)}+..........+\cfrac{1}{1+2+3+.....n)}=\cfrac{2n}{(n+1)}$

Let the given statement be

$P(n)$ i.e.,

$P(n)$ :
For

$n=1$ , we have

$P(1): 1=\cfrac{2\times 1}{1+1}=\cfrac{2}{2}=1$
which is true.
Let

$P(k)$ be true for some

$k\in N$ ,

$1+\cfrac{1}{1+2}+......\cfrac{1}{1+2+3}+......+\cfrac{1}{1+2+3+.....+k}=\cfrac{2k}{k+1}.........(i)$
We shall now prove that

$P(k+1)$ is true.
Consider

$1+\cfrac{1}{1+2}+\cfrac{1}{1+2+3}+.......+\cfrac{1}{1+2+3+....+k}+\cfrac{1}{1+2+3+......+k+(k+1)}$

$=\left( 1+\cfrac { 1 }{ 1+2 } +\cfrac { 1 }{ 1+2+3 } +.......+\cfrac { 1 }{ 1+2+3+....k } \right) +\left( \cfrac { 1 }{ 1+2+3+.......+k+(k+1) } \right)$

$=\cfrac { 2k }{ k+1 } +\cfrac { 1 }{ 1+2+3+.....+k+(k+1) }$ [Using

$(i)$ ]

$=\cfrac { 2k }{ k+1 } +\cfrac { 1 }{ \left( \cfrac { \left( k+1 \right) \left( k+1+1 \right) }{ 2 } \right) }$

$(\because 1+2+3+.....+n=\cfrac { n(n+1) }{ 2 } )$

$=\cfrac { 2k }{ \left( k+1 \right) } +\cfrac { 2 }{ \left( k+1 \right) \left( k+2 \right) }$

$=\cfrac { 2 }{ \left( k+1 \right) } \left( k+\cfrac { 1 }{ k+2 } \right)$

$=\dfrac{2(k^2+2k+1)}{(k+1)(k+2)}$

$=\dfrac{2(k+1)}{k+2}$
Hence, the given result is true for

$k+1$ .
So, by mathematical induction method, the given result is true for all

$n\in N$

Prove the following by using the principle of mathematical induction for all $n\in N: 1.2+2.3+3.4+......+n(n+1)=\left[ \cfrac { n(n+1)(n+2) }{ 3 } \right]$

Let the given statement be

$P(n)$ i.e.,

$1.2+2.3+3.4+......+n(n+1)=\left[ \cfrac { n(n+1)(n+2) }{ 3 } \right]$

$P(n)$ :
For

$n=1$ , we have

$1.2=2\cfrac{1(1+1)(1+2)}{3}=\cfrac{1.2.3}{3}=2$ , which is true.
Let

$P(k)$ be true for some positive integer

$k$ i.e.,

$1.2+2.3+3.4+.....+k.(k+1)=\left[ \cfrac { k(k+1)(k+2) }{ 3 } \right] ........(i)$
We shall now prove that

$P(k+1)$ is true.
Consider

$1.2+2.3+3.4+......+k.(k+1)+(k+1)+(k+1).(k+2)$

$=[1.2+2.3+3.4+.....+k.(k+1)] +(k+1).(k+2)$

$=\cfrac{k(k+1)(k+2)}{3}+(k+1)(k+2)$ [Using

$(i)$ ]

$=(k+1)(k+2)(\cfrac{k}{3}+1)$

$=\cfrac{(k+1)(k+2)(k+3)}{3}$

$=\cfrac{(k+1)(k+1+1)(k+1+2)}{3}$
Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$

Prove the following by using the principle of mathematical induction for all $n\in N$
${ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+.......+{ n }^{ 3 }={ \left[ \cfrac { n(n+1) }{ 2 } \right] }^{ 2 }$

Let the given statement be

$P(n)$ , i.e.,

${ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+.......+{ n }^{ 3 }={ \left( \cfrac { n(n+1) }{ 2 } \right) }^{ 2 }$

$P(n)$ :
For

$n=1$ , we have

$P(1):{1}^{3}=1={ \left( \cfrac { 1(1+1) }{ 2 } \right) }^{ 2 }\quad ={ \left( \cfrac { 1\times 2 }{ 2 } \right) }^{ 2 }={ 1 }^{ 2 }=1$ , which is true
Let

$P(k)$ be true for some positive integer

$k$ , i.e.,

${ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+.......+{ k }^{ 3 }={ \left( \cfrac { k(k+1) }{ 2 } \right) }^{ 2 }..........(i)$
We shall now prove that

$P(k+1)$ is true.
Consider

${ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+.......+{ k }^{ 3 }+{(k+1)}^{3}$

${ \left( \cfrac { k(k+1) }{ 2 } \right) }^{ 2 }+{ \left( k+1 \right) }^{ 3 }$ [Using

$(i)$ ]

$=\cfrac { { k }^{ 2 }{ \left( k+1 \right) }^{ 2 } }{ 4 } +{ \left( k+1 \right) }^{ 3 }$

$=\cfrac { { k }^{ 2 }{ \left( k+1 \right) }^{ 2 }+4{ \left( k+1 \right) }^{ 3 } }{ 4 } =\cfrac { { \left( k+1 \right) }^{ 2 }\left\{ { k }^{ 2 }+4(k+1) \right\} }{ 4 }$

$=\cfrac { { \left( k+1 \right) }^{ 2 }\left\{ { k }^{ 2 }+4k+4 \right\} }{ 4 } \quad =\cfrac { { \left( k+1 \right) }^{ 2 }{ \left( k+2 \right) }^{ 2 } }{ 4 }$

$=\cfrac { { \left( k+1 \right) }^{ 2 }{ \left( k+1+1 \right) }^{ 2 } }{ 4 }$

${ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+.......+{ k }^{ 3 }+{k+1}^{3}={ \left( \cfrac { \left( k+1 \right) \left( k+1+1 \right) }{ 2 } \right) }^{ 2 }$
Thus

$P(k+1)$ is true whenever

$P(k)$ is true
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$ .

Prove the following by using the principle of mathematical induction for all $n\in N: 1.2+2.{2}^{2}+3.{2}^{2}+.......+n.{2}^{n}=(n-1){2}^{n+1}+2$

Let the given statement be

$P(n)$ , i.e.,

$P(n): 1.2+2.{2}^{2}+3.{2}^{2}+........+n.{2}^{n}=(n-1){2}^{n+1}+2$
For

$n=1$ , we have

$P(1):1.2=2=(1-1){2}^{1+1}+2=0+2=2$ , which is true.
Let

$P(k)$ be true for some

$k\in N$ . i.e.,

$1.2+2.{2}^{2}+3.{2}^{2}+...... +k.{2}^{k}=(k-1){2}^{k+1}+2..........(i)$
We shall now prove that

$P(k+1)$ is true.
i.e.

$1.2+2.{ 2 }^{ 2 }+3.{ 2 }^{ 3 }+....k.{ 2 }^{ k } +(k+1).{ 2 }^{ k+1 }=k2^{k+2}+2$
Consider LHS,

$1.2+2.{ 2 }^{ 2 }+3.{ 2 }^{ 3 }+....k.{ 2 }^{ k } +(k+1).{ 2 }^{ k+1 }$

$=\left( k-1 \right) { 2 }^{ k+1 }+2+(k+1){ 2 }^{ k+1 }$

$={ 2 }^{ k+1 }\left\{ (k-1)+(k+1) \right\} +2$

$=k.{ 2 }^{ (k+1)}+2$

$=\left\{ (k+1)-1 \right\} { 2 }^{ (k+1)+1 }+2$
Thus

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$ .

Prove the following by using the principle of mathematical induction for all $n\in N: \cfrac{1}{2.5}+\cfrac{1}{5.8}+\cfrac{1}{8.11}+......+\cfrac{1}{(3n-1)(3n+2)}=\cfrac{n}{(6n+4)}$

Let the given statement be

$P(n)$ i.e

$P(n) : \cfrac{1}{2.5}+\cfrac{1}{5.8}+\cfrac{1}{8.11}+......+\cfrac{1}{(3n-1)(3n+2)}=\cfrac{n}{(6n+4)}$
For

$n=1$ , we have

$P(1)=\cfrac{1}{2.5}=\cfrac{1}{10}=\cfrac{1}{6.1+4}+\cfrac{1}{10}$ , which is true.
Let

$P(k)$ be true for some

$k\in N$ i.e.,

$\cfrac{1}{2.5}+\cfrac{1}{5.8}+\cfrac{1}{8.11}+......+\cfrac{1}{(3k-1)(3k+2)}=\cfrac{k}{6k+4}.......(i)$
We shall now prove that

$P(k+1)$ is true.
Consider

$\cfrac{1}{2.5}+\cfrac{1}{5.8}+\cfrac{1}{8.11}+.......+\cfrac{1}{(3k-1)(3k+2)}+\cfrac { 1 }{ \left\{ 3(k+1)-1 \right\} \left\{ 3(k+1)+2 \right\} }$

$=\cfrac { k }{ 6k+4 } +\cfrac { 1 }{ (3k+3-1)(3k+3+2 }$

$=\cfrac { k }{ 6k+4 } +\cfrac { 1 }{ (3k+2)(3k+5) }$

$=\cfrac { 1 }{ (3k+2) } \left( \cfrac { k }{ 2 } +\cfrac { 1 }{ 3k+5 } \right)$

$=\cfrac { 1 }{ (3k+2) } \left( \cfrac { k(3k+5)+2 }{ 2(3k+5) } \right)$

$=\cfrac { 1 }{ (3k+2) } \left( \cfrac { (3k+2)(k+1) }{ 2(3k+5) } \right)$

$=\cfrac { (k+1) }{ 6k+10 }$

$=\cfrac { (k+1) }{ 6(k+1)+4 }$
Thus

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$

Prove the following by using the principle of mathematical induction for all $n\in N: \left( 1+\cfrac { 3 }{ 1 } \right) \left( 1+\cfrac { 5 }{ 4 } \right) \left( 1+\cfrac { 7 }{ 9 } \right) ......\left( 1+\cfrac { (2n+1) }{ { n }^{ 2 } } \right) ={ (n+1) }^{ 2 }$

Let the given statement be

$P(n)$ , i.e.,

$P(n):\left( 1+\cfrac { 3 }{ 1 } \right) \left( 1+\cfrac { 5 }{ 4 } \right) \left( 1+\cfrac { 7 }{ 9 } \right) ......\left( 1+\cfrac { (2n+1) }{ { n }^{ 2 } } \right) ={ (n+1) }^{ 2 }$
For

$n=1$ , we have

$P(1): (1+\cfrac{3}{1})=4={(1+1)}^{2}={2}^{2}=4$ , which is true.
Let

$P(k)$ be true for some

$k\in N$ , i.e

$\left( 1+\cfrac { 3 }{ 1 } \right) \left( 1+\cfrac { 5 }{ 4 } \right) \left( 1+\cfrac { 7 }{ 9 } \right) ......\left( 1+\cfrac { (2k+1) }{ { k }^{ 2 } } \right) ={ (k+1) }^{ 2 }..............(1)$
We shall now prove that

$P(k+1)$ is true.
Consider

$\left( 1+\cfrac { 3 }{ 1 } \right) \left( 1+\cfrac { 5 }{ 4 } \right) \left( 1+\cfrac { 7 }{ 9 } \right) ......\left( 1+\cfrac { (2k+1) }{ { k }^{ 2 } } \right) \left( 1+\cfrac { \left\{ 2(k+1)+1 \right\} }{ { \left( k+1 \right) }^{ 2 } } \right)$

$={ \left( k+1 \right) }^{ 2 }\left[ 1+\cfrac { \left\{ 2(k+1)+1 \right\} }{ { \left( k+1 \right) }^{ 2 } } \right]$ [Using

$(i)$ ]

$={ \left( k+1 \right) }^{ 2 }\left[ \cfrac { { (k+1) }^{ 2 }+2(k+1)+1 }{ { (k+1) }^{ 2 } } \right]$

$={ (k+1) }^{ 2 }+2(k+1)+1$

$={ \left\{ (k+1)+1 \right\} }^{ 2 }$
Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$ .

Prove the following by using the principle of mathematical induction for all $n\in N:{10}^{2n-1}+1$ is divisible by $11$

Let the given statement be

$P(n)$ i.e.,

$P(n):{10}^{2n-1}+1$ is divisible by

$11$

$P(n)$ is true for

$n=1$ since

$P(1)={10}^{2.1-1}+1=11,$ which is divisible by

$1$

Let

$P(k)$ be true for some

$k\in N$ , i.e.,

${10}^{2k-1}+1=11m$ , where

$m\in N..........(i)$

We shall now prove that

$P(k+1)$ is true whenever

$P(k)$ is true.
Consider

${10}^{2(k+1)-1}+1$

$={10}^{2k+2-1}+1$

$={10}^{2}({10}^{2k-1})+1$

$={10}^{2}(11m-1)+1$ [Using

$(i)$ ]

$=100\times 11m-99$

$=11(100m-9)$

$=11r$ , where

$r=(100m-9)$ is some natural number
Therefore,

${10}^{2(k+1)-1}+1$ is divisible by

$11$ .
Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural number i.e,

$n$

Prove the following by using the principle of mathematical induction for all $n\in N:{x}^{2n}-{y}^{2n}$ is divisible by $x+y$ .

Let the given statement be

$P(n)$ i.e.

$P(n): {x}^{2n}-{y}^{2n}$ is divisible by

$x+y$ .

$P(n)$ is true for

$n=1$ since

${x}^{2\times 1}-{y}^{2\times 1}=(x+y)(x-y)$ is divisible by

$(x+y)$ .

Let

$P(k)$ be true for some positive integer

$k$ , i.e.

${x}^{2k}-{y}^{2k}$ is divisble by

$x+y$

${x}^{2k}-{y}^{2k}=m(x+y)$ , where

$m\in N......(i)$
We shall now prove that

$P(k+1)$ is true whenever

$P(k)$ is true.

Consider

${x}^{2(k+1)}-{y}^{2(k+1)}$

$={x}^{2k}.{x}^{2}-{y}^{2k}.{y}^{2}$

$={x}^{2}\left\{ m(x+y)+{ y }^{ 2k } \right\} -{ y }^{ 2k }.{ y }^{ 2 }$

$=m(x+y){ x }^{ 2 }+{ y }^{ 2k }({ x }^{ 2 }-{ y }^{ 2 })\quad$

$=m(x+y){ x }^{ 2 }+{ y }^{ 2k }(x+y)(x-y)$

$=(x+y)\left\{ m{ x }^{ 2 }+{ y }^{ 2k }(x-y) \right\}$
Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$ .

Prove the following b y using the principle of mathematical induction for all $n\in N:1+2+3+.....+n<\cfrac{1}{8}{(2n+1)}^{2}$

Let the given statement be

$P(n)$ , i.e.,

$P(n):1+2+3+.....+n<\cfrac{1}{8}{(2n+1)}^{2}$

$P(n)$ is true for

$n=1$ since

$1<\cfrac{1}{8}{(2.1)}^{2}=\cfrac{9}{8}$

Let

$P(k)$ be true

$P(n)$ be true for some

$k\in N$ is true i.e.

$P(k):1+2+3+.....+k<\cfrac{1}{8}{(2k+1)}^{2}$ ......(i)
Consider

$(1+2+......k)+(k+1)< \cfrac{1}{8}{(2k+1)}^{2}+(k+1)$ [Using

$(i)$ ]

$= \cfrac{1}{8}\{{(2k+1)}^{2}+8(k+1)\}$

$=\cfrac{1}{8}(4{k}^{2}+4k+1+8k+8)$

$=\cfrac { 1 }{ 8 } \left( 4{ k }^{ 2 }+12k+9 \right)$

$=\cfrac{1}{8}(2k+3)^{2}$

$=\cfrac{1}{8} { \left\{ 2(k+1)+1 \right\} }^{ 2 }$

$(1+2+3+.....k)+(k+1)< \cfrac{1}{8}{(2k+1)}^{2}$
Hence,
Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers

$n$

Prove the following by using principle of mathematical induction for all $n\in N: \cfrac { 1 }{ 3.5 } +\cfrac { 1 }{ 5.7 } +\cfrac { 1 }{ 7.9 } +.....+\cfrac { 1 }{ (2n+1)(2n+3) } =\cfrac { n }{ 3(2n+3) }$

Let the given statement be

$P(n)$ i.e.,

$P(n):\cfrac { 1 }{ 3.5 } +\cfrac { 1 }{ 5.7 } +\cfrac { 1 }{ 7.9 } +.....+\cfrac { 1 }{ (2n+1)(2n+3) } =\cfrac { n }{ 3(2n+3) }$

For

$n=1$ we have

$P(1):\cfrac { 1 }{ 3.5 } =\cfrac { 1 }{ 3(2.1+3) } =\cfrac { 1 }{ 3.5 }$ , which is true.

Let

$P(k)$ be true for some

$k\in N$ , i.e.,

$P(k):\cfrac { 1 }{ 3.5 } +\cfrac { 1 }{ 5.7 } +\cfrac { 1 }{ 7.9 } +.....+\cfrac { 1 }{ (2k+1)(2k+3) } =\cfrac { k }{ 3(2k+3) } ........(i)$
We shall now prove that

$P(k+1)$ is true.

Consider

$\cfrac { 1 }{ 3.5 } +\cfrac { 1 }{ 5.7 } +\cfrac { 1 }{ 7.9 } +.....+\cfrac { 1 }{ (2k+1)(2k+3) } +\cfrac { 1 }{ \left\{ 2(k+1)+1 \right\} \left\{ 2(k+1)+3 \right\} }$

$=\cfrac { k }{ 3(2k+3) } +\cfrac { 1 }{ (2k+3)(2k+5) }$ [Using

$(i)$ ]

$=\cfrac { 1 }{ (2k+3) } \left\{ \cfrac { k }{ 3 } +\cfrac { 1 }{ (2k+5) } \right\}$

$=\cfrac { 1 }{ (2k+3) } \left\{ \cfrac { k(2k+5)+3 }{ 3(2k+5) } \right\}$

$=\cfrac { 1 }{ (2k+3) } \left\{ \cfrac { 2{ k }^{ 2 }+5k+3 }{ 3(2k+5) } \right\}$

$=\cfrac { 1 }{ (2k+3) } \left\{ \cfrac { 2{ k }^{ 2 }+2k+3k+3 }{ 3(2k+5) } \right\}$

$=\cfrac { 1 }{ (2k+3) } \left\{ \cfrac { 2{ k }(k+1)+3(k+1) }{ 3(2k+5) } \right\}$

$=\cfrac { \left( k+1 \right) \left( 2k+3 \right) }{ 3\left( 2k+3 \right) \left( 2k+5 \right) }$

$=\cfrac { \left( k+1 \right) }{ 3\left\{ 2(k+1)+1 \right\} }$
Thus

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction statement

$P(n)$ is true for all natural numbers i.e.,

$n$ .

Prove the following by using the principle of mathematical induction for all $n\in N:\cfrac { 1 }{ 1.4 } +\cfrac { 1 }{ 4.7 } +\cfrac { 1 }{ 7.10 } +.....+\cfrac { 1 }{ (3n-2)(3n+1) } =\cfrac { n }{ (3n+1) }$

Let the given statement be

$P(n)$ i.e.,

$P(n):\cfrac { 1 }{ 1.4 } +\cfrac { 1 }{ 4.7 } +\cfrac { 1 }{ 7.10 } +.....+\cfrac { 1 }{ (3n-2)(3n+1) } =\cfrac { n }{ (3n+1) }$
For

$n=1$ , we have

$P(1)=\cfrac { 1 }{ 1.4 } =\cfrac { 1 }{ 3.1+1 } =\cfrac { 1 }{ 4 } =\cfrac { 1 }{ 1.4 }$ , which is true.
Let

$P(k)$ be true for some

$k\in N$ , i.e.,

$P(k)=\cfrac { 1 }{ 1.4 } +\cfrac { 1 }{ 4.7 } +\cfrac { 1 }{ 7.10 } +.....+\cfrac { 1 }{ (3k-2)(3k+1) } =\cfrac { k }{ (3k+1) } ......(i)$
We shall now prove that

$P(k+1)$ is true.
Consider

$\cfrac { 1 }{ 1.4 } +\cfrac { 1 }{ 4.7 } +\cfrac { 1 }{ 7.10 } +.....+\cfrac { 1 }{ (3k-2)(3k+1) } +\cfrac { 1 }{ \left\{ 3(k+1)-2 \right\} \left\{ 3(k+1)+1 \right\} }$ [Using

$(i)$ ]

$=\cfrac { k }{ 3k+1 } +\cfrac { 1 }{ (3k+1)(3k+4) }$

$=\cfrac { 1 }{ (3k+1) } \left\{ k+\cfrac { 1 }{ (3k+4) } \right\}$

$=\cfrac { 1 }{ (3k+1) } \left\{ \cfrac { k(3k+4)+1 }{ (3k+4) } \right\}$

$=\cfrac { 1 }{ (3k+1) } \left\{ \cfrac { 3{ k }^{ 2 }+4k+1 }{ (3k+4) } \right\}$

$=\cfrac { \left( 3k+1 \right) \left( k+1 \right) }{ \left( 3k+1 \right) \left( 3k+4 \right) }$

$=\cfrac { \left( k+1 \right) }{ 3(k+1)+1 }$
Thus

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$ .

Prove the following by using the principle of mathematical induction for all $n\in N: P(n):a+ar+a{r}^{2}+......+a{r}^{n-1}=\cfrac{a({r}^{n}-1)}{r-1}$

Let the given statement be

$P(n)$ i.e.,

$P(n):a+ar+a{r}^{2}+......+a{r}^{n-1}=\cfrac{a({r}^{n}-1)}{r-1}$
For

$n=1$ , we have

$P(1):a=\cfrac{a({r}^{1}-1)}{(r-1)}=a$ , which is true.
Let

$P(k)$ be true for some

$k\in N$ , i.e

$a+ar+a{r}^{2}+.......+a{r}^{k-1}=\cfrac{a({r}^{k}-1)}{r-1}...........(i)$
We shall now prove that

$P(k+1)$ is true.
Consider

$a+ar+a{r}^{2}+...........a{r}^{k-1}-1$

$=\cfrac{a({r}^{k}-1)}{r-1}+a{r}^{k}$ [Using (i)]

$=\cfrac { a\left( { r }^{ k }-1 \right) +a{ r }^{ k+1 }-a{ r }^{ k } }{ r-1 } \quad$

$=\cfrac { a{ r }^{ k }-a+a{ r }^{ k+1 }-a{ r }^{ k } }{ r-1 }$

$=\cfrac { a{ r }^{ k+1 }-a }{ r-1 }$

$=\cfrac { a({ r }^{ k+1 }-1) }{ r-1 }$
Thus

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$

Prove the following by using the principle of mathematical induction for all $n\in N:{ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+.......+({ 2n-1) }^{ 2 }=\cfrac { n(2n-1)(2n+1) }{ 3 }$

Let the given statement be

$P(n)$ i.e.,

${ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+.......+({ 2n-1) }^{ 2 }=\cfrac { n(2n-1)(2n+1) }{ 3 }$
For

$n=1$ we have

$P(1)={ 1 }^{ 2 }=1=\cfrac { 1(2.1-1)(2.1+1) }{ 3 } =\cfrac { 1.1.3 }{ 3 } =1$ , which is true.
Let

$P(k)$ be true for some

$k\in N$ i.e.,

${ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+.......+({ 2n-1) }^{ 2 }=\cfrac { k(2k-1)(2k+1) }{ 3 } .....(i)$
We shall now prove that

$P(k+1)$ is true.
Consider

${ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+.......+({ 2k-1) }^{ 2 } +{ \left\{ 2(k+1)-1 \right\} }^{ 2 }$ [Using

$(i)$ ]

$=\cfrac { k(2k-1)(2k+1) }{ 3 } +{ \left( 2k+2-1 \right) }^{ 2 }$

$=\cfrac { k(2k-1)(2k+1) }{ 3 } +{ \left( 2k+1 \right) }^{ 2 }$

$=\cfrac { k(2k-1)(2k+1)+3{ \left( 2k+1 \right) }^{ 2 } }{ 3 }$

$=\cfrac { (2k+1)\left\{ k(2k-1)+3(2k+1) \right\} }{ 3 }$

$=\cfrac { (2k+1)( 2{ k }^{ 2 }-k+6k+3 ) }{ 3 }$

$=\cfrac { (2k+1)( 2{ k }^{ 2 }+5k+3 ) }{ 3 }$

$=\cfrac { (2k+1)(2{ k }^{ 2 }+2k+3k+3 ) }{ 3 }$

$=\cfrac { (2k+1)\left\{ 2k(k+1)+3(k+1) \right\} }{ 3 }$

$=\cfrac { (2k+1)(k+1)(2k+3) }{ 3 }$

$=\cfrac { (k+1)\left\{ 2(k+1)-1 \right\} \left\{ 2(k+1)+1 \right\} }{ 3 }$
Thus

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$ .

Prove the following by using the principle of mathematical induction for all $n\in N:n(n+1)(n+5)$ is multiple of $3$ .

Let the given statement be

$P(n)$ , i.e.,

$P(n):n(n+1)(n+5)$ , which is a multiple of

$3$ .

$P(n)$ is true for

$n=1$ since

$1(1+1)(1+5)=12$ , which is a multiple of

$3$ .

Let

$P(k)$ be true for some

$k\in N$ i.e.,

$k(k+1)(k+5)$ is a multiplie of

$3$

$k(k+1)(k+5)=3m$ , where

$m\in N.......(1)$

We shall now prove that

$P(k+1)$ is true whenever

$P(k)$ is true.
Consider

$(k+1)\left\{ (k+1)+1 \right\} \left\{ (k+1)+5 \right\}$

$=(k+1)(k+2)\left\{ (k+5)+1 \right\}$

$=(k+1)(k+2)(k+5)+(k+1)(k+2)$

$=k(k+1)(k+5)+2(k+1)(k+5) +(k+1)(k+2)$

$=3m+(k+1)\left\{ 2(k+5)+(k+2) \right\}$

$=3m+(k+1)(2k+10+k+2 )$

$=3m+(k+1)(3k+12)$

$=3m+3(k+1)(k+4)$

$=3\left\{ m+(k+1)(k+4 \right\} =3\times q$
where

$q=\left\{ m+(k+1)(k+4) \right\}$
is some natural number
Therefore

$(k+1)\left\{ (k+1)+1 \right\} \left\{ (k+1)+5 \right\}$ is a multiple of

$3$ .
Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$ .

Prove the following by using the principle of mathematical induction for all $n\in N:\left( 1+\cfrac { 1 }{ 1 } \right) \left( 1+\cfrac { 1 }{ 2 } \right) \left( 1+\cfrac { 1 }{ 3 } \right) ......\left( 1+\cfrac { 1 }{ n } \right) =(n+1)\quad$

Let the given statement be

$P(n)$ i.e.,

$P(n):\left( 1+\cfrac { 1 }{ 1 } \right) \left( 1+\cfrac { 1 }{ 2 } \right) \left( 1+\cfrac { 1 }{ 3 } \right) ......\left( 1+\cfrac { 1 }{ n } \right) =(n+1)\quad$
For

$n=1$ , we have

$P(1):\left( 1+\cfrac { 1 }{ 1 } \right) =2=(1+1)$ , which is true.

Let

$P(k)$ is true for some

$k\in N$

$\left( 1+\cfrac { 1 }{ 1 } \right) \left( 1+\cfrac { 1 }{ 2 } \right) \left( 1+\cfrac { 1 }{ 3 } \right) ......\left( 1+\cfrac { 1 }{ k } \right) =(k+1).........(i)$
We shall now prove that

$P(k+1)$ is true.
Consider

$\left[ \left( 1+\cfrac { 1 }{ 1 } \right) \left( 1+\cfrac { 1 }{ 2 } \right) \left( 1+\cfrac { 1 }{ 3 } \right) ......\left( 1+\cfrac { 1 }{ k } \right) \right] \left( 1+\cfrac { 1 }{ k+1 } \right)$

$=(k+1)\left( 1+\cfrac { 1 }{ k+1 } \right)$ [Using

$(i)$ ]

$=(k+1)\left( \cfrac { (k+1)+1 }{ k+1 } \right)$

$=(k+1)+1$
Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$

Find counter examples to disprove the following statement.
A quadrilateral with all sides are equal is a square.

Given statement:

A quadrilateral with all sides are equal is a square.

But, also a rhombus has all sides equal.

A rhombus in which all angles are

$90°$ is a square.

Hence, the required example is a rhombus.

Using mathematical induction, prove that $\cfrac{d}{dx}({x}^{n})=n{x}^{n-1}$ for all positive integers $n$ .

Check the given expression for

$n=1$

L.H.S.

$= \cfrac{d}{dx}(x^1) =\cfrac{dx}{dx} = 1$

R.H.S.

$= 1.x^{1-1} = 1$

Hence, it is true for

$n=1$

Let the expression be true for

$n=k$

$\cfrac{d}{dx}(x^k) = kx^{k-1}$

Now, check for

$n = k+1$

L.H.S.

$=\cfrac{d}{dx}(x^{k+1}) = \cfrac{d}{dx}(x^k.x) = x\cfrac{d}{dx}(x^k) + x^k\cfrac{d}{dx}(x)\\=x(kx^{k-1}) + x^k(1)=kx^k+x^k\\=(k+1)x^k = (k+1)x^{(k+1) - 1}$

Hence, the given expression is true for

$n = k+1$ too.

By principle of mathematical induction,

$\cfrac{d}{dx}(x^n) = nx^{n-1}$

Prove the following by using the principle of mathematical induction for all $n\in N:{41}^{n}-{14}^{n}$ is a multiple of $27$ .

Let the given statement be

$P(n)$ . i.e.,

$P(n):{41}^{n}-{14}^{n}$ is a multiple of

$27$

$P(n)$ is true for

$n=1$ since

${41}^{1}-{14}^{1}=27$ , which is a multiple of

$27$

Let

$P(k)$ be true for some positive integer

$k$ , i.e.,

${41}^{k}-{14}^{k}$ is a multiple of

$27$

${41}^{k}-{14}^{k}=27m$ , where

$m\in N.........(i)$
We shall now prove that

$P(k+1)$ is true whenever

$P(k)$ is true.

Consider

${41}^{k+1}-{14}^{k+1}$

$={41}^{k}.41-{14}^{k}.14$

$=41(27m+{ 14 }^{ k })-{ 14 }^{ k }14$ (using(i))

$=41.27m+41.{14}^{k}-14.{14}^{k}$

$=41.27m+27.{14}^{k}$

$=27(41m+{14}^{k})$

$=27\times r$ , where

$r=(41m+{14}^{k})$ is a natural number
Therefore,

${41}^{k+1}-{14}^{k+1}$ is a multiple if

$27$ .
Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$ .

Prove the following by using the principle of mathematical induction for all $n\in N:(2n+7)< {(n+3)}^{2}$ .

Let the given statement be

$P(n)$ , i.e.,

$P(n):(2n+7)< {(n+3)}^{2}$

$P(n)$ is true for

$n=1$ since

$2.1+7=9< {(1+3)}^{2}=16$ , which is true.

Let

$P(k)$ be true for some positive integer

$k$ , i.e.,

$(2k+7)< {(k+3)}^{2}........(i)$
We shall now prove that

$P(k+1)$ is true whenever

$P(k)$ is true.

Consider

$2(k+1)+7 =\{(2k+7)+2\} < { \left( k+3 \right) }^{ 2 }+2$ [Using

$(i)$ ]

$2(k+1)+7< {k}^{2}+6k+9+2$

$2(k+1)+7< {k}^{2}+6k+11$
Now,

${k}^{2}+6k+11< {k}^{2}+8k+16$

$\therefore$

$2(k+1)+7< {(k+4)}^{2}$

$2(k+1)+7< { \left\{ (k+1)+3 \right\} }^{ 2 }$
Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$ .

Prove the following by using the principle of mathematical induction for all $n\in N:{3}^{2n+2}-8n-9$ is divisible by $8$ .

Let the given statement be

$P(n)$ , i.e.,

$P(n):{3}^{2n+2}-8n-9$ is divisible by

$8$ .

$P(n)$ is true for

$n=1$ since

${3}^{2\times 1+2}-8\times 1-9=64$ , which is divisible by

$8$ .

Let

$P(k)$ be true for some

$k\in N$ , i.e.,

${3}^{2k+2}-8k-9$ is divisible by

$8$

${3}^{2k+2}-8k-9=8m$ ; where

$m\in N........(i)$
We shall now prove that

$P(k+1)$ is true whenever

$P(k)$ is true.

Consider

${3}^{2(k+1)+2}-8(k+1)-9$

$={3}^{2k+2}.{3}^{2}-8k-8-9$

$={3}^{2}({3}^{3k+2}-8k-9+8k+9)-8k-17$

$={3}^{2}({3}^{2k+2}-8k-9)+{3}^{2}(8k+9)-8k-17$

$=9.8m+9(8k+9)-8k-17$

$=9.8m+72k+81-8k-17$

$=9.8m+64k+64$

$=8(9m+8k+8)$

$=8r$ , where

$r=(9m+8k+8)$ is a natural number
Therefore,

${3}^{2(k+1)+2}-8(k+1)-9$ is divisible by

$8$ .
Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.
Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n$

Prove by induction:
${1}^{2}+{2}^{2}+{3}^{2}+........+{n}^{2}=\cfrac{1}{6}n(n+1)(2n+1)$

The statement to be proved is:

$P(n):1^2+2^2+..+n^2=\dfrac{1}{6}n(n+1)(2n+1)$

Step 1: Verify that

$P(n)$ is true for

$n=1$

$P(1):\dfrac{1}{6}1(1+1)(2+1)$

$P(1):1=1$

Therefore,

$P(1)$ is true.

Step 2: Assume that the statement is true for

$n=k$

Let us assume that the below statement is true:

$P(k):1^2+2^2+..+k^2=\dfrac{1}{6}k(k+1)(2k+1)$

Step 3: Verify that the statement is true for

$n=k+1$

We need to prove that:

$1^2+2^2+..+{(k+1)}^2=\dfrac{1}{6}(k+1)(k+2)(2k+3)$

$LHS=1^2+2^2+...+(k+1)^2$

$=\dfrac{1}{6}k(k+1)(2k+1)+(k+1)^2$

$=(k+1)\left ( \dfrac{1}{6}k(2k+1)+(k+1) \right )$

$=(k+1)\left ( \dfrac{k(2k+1)+6(k+1)}{6} \right )$

$=\dfrac{k+1}{6}(2k^2+7k+6)$

$=\dfrac{k+1}{6}(2k^2+3k+4k+6)$

$=\dfrac{k+1}{6}(k(2k+3)+2(2k+3))$

$=\dfrac{1}{6}(k+1)(k+2)(2k+3)$

$=RHS$

Hence,

$P(n)$ is true for all values of

$n$ by principle of mathematical induction.

Prove by induction:
${x}^{n}-{y}^{n}$ is divisible by $x+y$ when $n$ is even.

The statement to be proved is:

$P(n):x^n-y^n$ is divisble by

$x+y$ where

$n$ is even.

Step 1: Verify that the statement is true for the smallest value of

$n$ , here,

$n=2$

$P(2):x^2-y^2$ is divisible by

$x+y$

$P(2):(x+y)(x-y)$ is divisible by

$x+y$ , which is true.

Therefore

$P(2)$ is true.

Step 2: Assume that the statement is true for

$k$

Let us assume that

$P(k): x^k-y^k$ is divisible by

$x+y$ where

$k$ is even.

Step 3: Verify that the statement is true for the next possible integer, here for

$n=k+2$

$x^{k+2}-y^{k+2}=x^{k+2}-x^2y^k+x^2y^k-y^{k+2}$

$=x^2(x^k-y^k)+y^k(x^2-y^2)$

Since

$(x^k-y^k)$ and

$(x^2-y^2)$ are both divisible by

$(x+y)$ , the complete equality is divisible by

$x+y$

Therefore,

$P(k+2): x^{k+2}-y^{k+2}$ is divisible by

$x+y$ where

$k+2$ is even.

Therefore by principle of mathematical induction,

$P(n)$ is true.

Show that ${ 2 }^{ 2n }-3n-1$ is divisible by $9$ for all positive integral values of $n$ .

We will prove this by induction.

Let

$I=2^{2n}-3n-1=4^n-3n-1$

For

$n=1$ ,

$I=4-3-1=0$ and zero is divisible by 9.

Now, let us assume

$I$ to be divisible by

$9$ for some

$n=k$ .

That is

$I=4^k-3k-1=9\times r$ ------(1) (where r is any integer)

We need to prove that

$I$ is divisible by

$9$ for

$n=k+1$

$I=4^{k+1}-3(k+1)-1=4\times 4^k-3k-4$

$=4(4^k-1)-3k$

Replace

$4^k-1=9r+3k$ ----by using eq. (1),

We get

$I=4(9r+3k)-3k=36r+12k-3k=36r+9k=9(4r+1)$

$r$ is an integer, therefore

$4r+1$ is also an integer and hence

$9(4r+1)$ is also an integer.

Hence for

$n=k+1$ ,

$I$ is divisible by

$9$ .

Hence by induction, we have proved that

$2^{2n}-3n-1$ is divisible by 9.

Prove by induction:
$2+{2}^{2}+{2}^{3}+........+{2}^{n}=2({2}^{n}-1)$

The statement to be proved is:

$P(n):2+2^2+2^3+...+2^n=2(2^n-1)$

Step 1: Prove that the statement is true for

$n=1$

$P(1):2^1=2(2^1-1)$

$P(1):2=2$

Hence, the statement is true for

$n=1$

Step 2: Assume that the statement is true for

$n=k$

Let us assume that the below statement is true:

$P(k):2+2^2+...+2^k=2(2^k-1)$

Step 3: Prove that the statement is true for

$n=k+1$

We need to prove that:

$2+2^2...+2^{k+1}=2(2^{k+1}-1)$

$LHS=2+2^{2}+...+2^{k}+2^{k+1}$

$=2(2^k-1)+2^{k+1}$

$=2(2^k-1+2^k)$

$=2(2.2^k-1)$

$=2(2^{k+1}-1)$

$=RHS$

Therefore,

$P(n)$ is true for all values of

$n$ by principle of mathematical induction.

Prove the following by the principle of mathematical induction: $(n\epsilon N)$
$1^{2} . 2 + 2^{2} . 3 + .... + n^{2} (n + 1) = \dfrac {n(n + 1)(n + 2)(3n + 1)}{12}$ .

1167559_879092_ans_e7de87f83258447194a9232783821457.jpg

Prove the following by the principle of Mathematical Induction.
$(2n + 1)(2n - 1)$ is an odd number for all $n\epsilon N$ .

1160524_879072_ans_f47fe691795846a5aecb785b166b1ea1.jpg

Prove by induction:
$\cfrac{1}{1.2}+\cfrac{1}{2.3}+\cfrac{1}{3.4}+.....$ to $n$ terms $=\cfrac{n}{n+1}$

The statement to be proved is:

$P(n):\dfrac{1}{1.2}+\dfrac{1}{2.3}+...$ to

$n$ terms

$=\dfrac{n}{n+1}$

Step 1: Prove that the statement is true for

$n=1$

$P(1):\dfrac{1}{1.2}=\dfrac{1}{1+1}$

$P(1):\dfrac{1}{2}=\dfrac{1}{2}$

Hence, the statement is true for

$n=1$

Step 2: Assume that the statement is true for

$n=k$

Let us assume that the below statement is true:

$P(k):\dfrac{1}{1.2}+\dfrac{1}{2.3}+... +\dfrac{1}{k(k+1)}=\dfrac{k}{k+1}$

Step 3: Prove that the statement is true for

$n=k+1$

We need to prove that:

$P(k+1):\dfrac{1}{1.2}+\dfrac{1}{2.3}+... +\dfrac{1}{(k+1)(k+2)}=\dfrac{k+1}{k+2}$

$LHS=\dfrac{1}{1.2}+\dfrac{1}{2.3}+... +\dfrac{1}{(k+1)(k+2)}$

$=\dfrac{k}{k+1}+\dfrac{1}{(k+1)(k+2)}$

$=\dfrac{1}{k+1}\left ( k+\dfrac{1}{(k+2)} \right )$

$=\dfrac{1}{k+1}\left ( \dfrac{k(k+2)+1}{(k+2)} \right )$

$=\dfrac{1}{k+1}\left ( \dfrac{k^2+2k+1}{(k+2)} \right )$

$=\dfrac{1}{k+1}\left ( \dfrac{(k+1)^2}{(k+2)} \right )$

$=\dfrac{k+1}{k+2}$

$=RHS$

Therefore, the statement is true for all

$n$ by principle of mathematical induction.

Prove by induction:
$1+3+5+.....+(2n-1)={n}^{2}$

The statement to be proved is:

$P(n)=1+3+5+...+(2n-1)=n^2$

Step 1: Veriy that

$P(1)$ is correct:

$P(1):{2(1)-1}=1^2 \\ P(1): 1=1$

Therefore it is verified that

$P(1)$ is correct.

Step 2: Assume that

$P(k)$ is true

Let us assume that:

$P(k): 1+3+5+...+(2k-1)=k^2$ holds.

Step 3: Prove that

$P(k+1)$ is true.

$LHS=1+3+5+...+(2(k+1)-1)$

$=P(k)+(2k+1)$

$=k^2 + 2k + 1$

$=(k+1)^2$

$=RHS$

Hence,

$P(n)$ is true by Principle of Mathematical Induction

If $p$ and $q$ are any two positive integers, show that $|\underline {pq}$ is divisible by $(|\underline {p})^{q} . |\underline {q}$ and by $(|\underline {q})^{p} . |\underline {p.}$

These results are easily established by Induction.

Suppose that,

$|\underline { p\left( q-1 \right) }$ is divisible by

${ \left( |\underline { p } \right) }^{ q-1 }|\underline { q-1 } .$

Now

$\cfrac { \left( pq \right) ! }{ { \left( p! \right) }^{ q }q! } \div \cfrac { \left( p\left( q-1 \right) \right) }{ { \left( p! \right) }^{ q-1 }\left( q-1 \right) ! } =\cfrac { \left( pq \right) ! }{ \left( pq-p \right) ! } \times \cfrac { \left( q-1 \right) ! }{ p!q! }$

$=\cfrac { pq\left( pq-1 \right) \left( pq-2 \right) ......pfactors }{ pq\left( p-1 \right) ! }$

$=\cfrac { \left( pq-1 \right) \left( pq-2 \right) .......\left( p-1 \right) factors }{ \left( p-1 \right) ! }$

But

$\cfrac { p! }{ p!1! }$ is an integer.

Hence,

$\cfrac { \left( 2p \right) ! }{ { \left( p! \right) }^{ 2 }2! }$ is an integer; and so on.

$(1 + x)^n >1$ + nx for n > 2 , n $\epsilon$ N
x > -1 x $\neq$ 0

Let P(n) be true for

$n \, = \, m$
Also when

$n \, =\, 1, \, (1+x)^1 \, = \, 1 \, + \, , \, x$ . so in this case, inequality does not hold.
But when

$n \, = \, 2$

$\therefore \, (1+x)^2 \, = \, 1 \, + \, 2x \, + \, x^2 \, > \, 1 \, + \, 2x$
Thus P(n) is true for

$n \, = \, 2$

$(1+m)^m > 1 \, + \, mx$ ....(1)
Since

$x>-1$ , multiplying by

$x \, + \, 1$ ,

$P(m+1) \, = \, (1+x)^{m+1} \, = \, (1+x)^{m}\, (1+x) > (1+mx)\,(1+x), \, \, by \,\,(1)$

$= \, 1 \, + \, (m+1)\, x \, + \, mx^2$

$= \, 1 \, + \, (m+1)\, x \, + \, mx^2> 1 \, + \, (m+1)x \,\, \because \, mx^2>0$
Above relation shows that P(n) is true for

$n \, = \, m \, + \, 1$
Hence P(n) is true universally if

$n \, \geq \, 2$

Prove by Mathematical induction that
${ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }...{ \left( 2n-1 \right) }^{ 2 }=\cfrac { n\left( 2n-1 \right) \left( 2n+1 \right) }{ 3 } \forall n\in N$

TO PROVE:

${ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }...+{ \left( 2n-1 \right) }^{ 2 }=\dfrac { n\left( 2n-1 \right) \left( 2n+1 \right) }{ 3 } ∀n\in N$

PROOF:

${ P\left( n \right) =1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }...+{ \left( 2n-1 \right) }^{ 2 }=\dfrac { n\left( 2n-1 \right) \left( 2n+1 \right) }{ 3 }$

$P\left( 1 \right) :{ \left( 2\times 1-1 \right) }^{ 2 }=\dfrac { 1\left( 2-1 \right) \left( 2+1 \right) }{ 3 }$

$\Rightarrow { \left( 1 \right) }^{ 2 }=1=\dfrac { 1\times 1\times 3 }{ 3 } =1$

$\therefore$ L.H.S=R.H.S (Proved)

$\therefore P(1)$ is true.

Now, let

$P(m)$ is true.

Then,

${ P\left( m \right) =1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }...+{ \left( 2m-1 \right) }^{ 2 }=\dfrac { m\left( 2m-1 \right) \left( 2m+1 \right) }{ 3 }$

Now, we have to prove that

$P(m+1)$ is also true.

${ P\left( m+1 \right) =1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }...+{ \left( 2m-1 \right) }^{ 2 }+{ \left[ 2\left( m+1 \right) -1 \right] }^{ 2 }$

${ = }P\left( m \right) +{ \left( 2m+2-1 \right) }^{ 2 }$

${ = }P\left( m \right) +{ \left( 2m+1 \right) }^{ 2 }$

$=\dfrac { m\left( 2m-1 \right) \left( 2m+1 \right) }{ 3 } +{ \left( 2m+1 \right) }^{ 2 }$

$=\dfrac { m\left( 2m-1 \right) \left( 2m+1 \right) +3{ \left( 2m+1 \right) }^{ 2 } }{ 3 }$

$=\dfrac { \left( 2m+1 \right) \left[ m\left( 2m-1 \right) +3{ \left( 2m+1 \right) } \right] }{ 3 }$

$=\dfrac { \left( 2m+1 \right) \left[ 2{ m }^{ 2 }-m+6m+3 \right] }{ 3 }$

$=\dfrac { \left( 2m+1 \right) \left[ 2{ m }^{ 2 }+5m+3 \right] }{ 3 }$

$=\dfrac { \left( 2m+1 \right) \left[ 2{ m }^{ 2 }+2m+3m+3 \right] }{ 3 }$

$=\dfrac { \left( 2m+1 \right) \left[ 2{ m }\left( m+1 \right) +3\left( m+1 \right) \right] }{ 3 }$

$=\dfrac { \left( 2m+1 \right) \left( 2m+3 \right) \left( m+1 \right) }{ 3 }$

$=\dfrac { \left( m+1 \right) \left[ 2\left( m+1 \right) +1 \right] \left[ 2\left( m+1 \right) -1 \right] }{ 3 }$

$\therefore p(m+1)$ is also true (Proved)

Prove $2 + 5 + 8 + 11 + ... + (3n - 1) =$ $\dfrac{1}{2}$ $n (3n + 1) n$ $\epsilon$ N

Let P(n) be true for

$n = m$ , that is, we suppose that

$P(m) = 2 + 5 + 8 + 11 + ... + (3m - 1) =$

$\dfrac{1}{2}$ m (3m + 1)
Now P(m + 1) = P(m) +

$T_{m \, + \,1}$

$= \, \dfrac{1}{2}m (3m \, + \, 1) \,+ \, [3(m \, + \, 1) \, - \, 1]$

$= \, \dfrac{1}{2}[3m^2 \, + \, m \, + \, 6m \, + \, 6 \, - \, 2]$

$= \, \dfrac{1}{2}[3m^2 \, + \, 7m \, + \, 4]$

$= \, \dfrac{1}{2}(m \, + \, 1)(3m \, + \, 4)$

$= \, \dfrac{1}{2}(m \, + \, 1)[3(m) \, + \, 1]$
Above relation shows that P(n) is true for n = m + 1.
Also when n = 1,P(n) = 2 =

$\dfrac{1}{2}(3 \, \cdot \, 1 \, + \, 1)$
n = 2, (P(n) = 2 + 5 = 7 =

$\dfrac{1}{2} \cdot 2 (3 \cdot 2 \, + \, 1)$
Above relation shows that P(n) is true for n = 1,2 etc.
Hence P(n) is universally true by mathematical induction.

Prove that $11^{n +2} + 12^{2n \, + \, 1}$ is divisible by $133$ for any non-negative integral $n$ .

First, prove it for the base case of

$n = 1$ :

$11^{(1+2)} + 12^{(2+1)}$
=

$11^3 + 12^3$
=

$1331 + 1728$
=

$133(10) + 1 + 1728$
=

$133(10) + 1729$
=

$133(10) + 133(13)$
=

$133(26)$

Assume it is true for a natural number

$k$ . Prove it is true for the number

$k+1$ :

True for

$k$ :

$11^{(k+2)} + 12^{(2k+1)} = 133(m)$ , where

$m$ is an integer.

$\text{For} k+1:$

$11^{k+1+2} + 12^{2(k+1)+1}$
=

$11^{k+2} \times11 + 12^{2k + 2 + 1}$
=

$11^(k+2) \times 11 + 12^(2k + 1) \times 12^2$
=

$11 \times 11^{k+2} + (11 + 133) \times 12^{2k+1}$
=

$11( 11^{k+2} + 12^{2k+1} ) + 133 \times 12^{2k+1}$
=

$11( 133m ) + 133 \times 12^{2k+1}$
=

$133 ( 11m + 12^{(2k + 1)})$ hence it is divisible by

$133$ .

Prove $\dfrac {1}{1 \cdot 2\cdot 3 } \, + \, \dfrac {1}{2 \cdot 3\cdot 4 } \, + \, \dfrac {1}{n (n + 1)(n + 2)} \, = \, \dfrac{1(n + 3)}{4(n + 1)(n + 2)}$

p(1) =

$\dfrac{1}{6} \, = \, \dfrac{1 \, \cdot 4}{4 (2 \, \cdot \, 4)} \, = \, \dfrac {1}{5}$
Similary p (2) also holds
Assums p(n) also holds

$\therefore p \, ( n \, + \, 1) \, = \, p(n ) \, + \, \dfrac{1}{ (n \, + 1)(n \, + \, 2)( n \, + \, 3)}$

$= \dfrac{n ( n \, + \, 3)} { 4 ( n + \, 1) ( n \, + \, 2)} \, + \, \dfrac{1}{ (n \, + 1)(n \, + \, 2)( n \, + \, 3)}$

$= \dfrac{1}{ 4(n \, + 1)(n \, + \, 2)( n \, + \, 3)} [ n (n \, + \, 3)^2 \,+ 4]$

$\dfrac {n^3 \, + \, 6n^2 + \, 9n \,+ 4 }{4 ( n + \, 1) ( n \, + \, 2)(n \, + \, 3)}$

$= \, \dfrac{(n \, + \, 1)^2 ( n + 4 ) }{4 ( n + \, 1) ( n \, + \, 2)(n \, + \, 3)}$

$= \dfrac{(n \, + \, 1)(n \, + \, 4) }{4 ( n + \, 1) ( n \, + \, 2)}$

$= \dfrac{N \, ( N \, + \, 3)}{4(N \, + \, 1)( M \, + \, 2) }$
Where N = n + 1

Prove the mathmatical induction that $\dfrac{1}{\log_x2 , \log_x4 }\, + \, \dfrac{1}{\log_x4 , \log_x8 } \, + \, ..... + \,\dfrac{1}{\log_x2^{n - 1} , \log_x4^n }$
= $\left (1- \dfrac{1}{n} \right ) \dfrac{1}{(\log_x2)^2} , x > 0 , \neq -1 n \epsilon N$

P(1) holds good as R.H.S. = 0 and L.H.S. = 0

$P(2) \, = \, \dfrac{1}{log_x2, \, log_x2^2} \, = \, \dfrac{1}{2(log_x2)^2}$

$= \,\left ( 1 \, - \, \dfrac{1}{2} \right )\, \dfrac{1}{(log_x2)^2}$
Assume

$P(n) \, = \,\left ( 1 \, - \, \dfrac{1}{n} \right )\, \dfrac{1}{(log_x2)^2}$ ....(1)

$P \, (n+1) \, = \, P(n) \, + \, \dfrac{1}{log_x2^n\cdot log_x2^{n+1}}$
Apply

$log \, P^q \, = \, q \, log \, p$ on 2nd term

$= \, \,\left ( 1 \, - \, \dfrac{1}{n} \right )\, \dfrac{1}{(log_x2)^2} \, + \, \dfrac{1}{1(n+1)} \cdot \dfrac{1}{(log_x2)^2} \, \, by \, \, (1)$

$= \, \, \left [ 1 \, - \, \dfrac{1}{n} \, + \,\left ( \dfrac{1}{n} \, - \, \frac{1}{n+1} \right ) \right ] \, \dfrac{1}{(log_x2)^2}$

$= \, \left (1 \, - \, \dfrac{1}{n+1} \right ) \dfrac{1}{(log_x2)^2}$

$\therefore \, P \,(n+1)$ holds good.
Hence the result is true universally.

$\sqrt 2 + \sqrt 2 + \sqrt 2 + ...+ n\ terms = 2\cos \dfrac{\pi}{2^{n +1}} , n \epsilon N$

Using Mathematical inducion

$Step (I) = n=1$

$P(1) \, = \, \sqrt{2} \, = \, 2\,cos\,\dfrac{\pi}{2^2} \, = \, 2\, \cdot \, \dfrac{1}{\sqrt{2}} \, = \, \sqrt{2}$

$\therefore P(1)$ holds .

$step( II)$ = Assume that

$P(n)$ is true

$P(n)=2 cos\dfrac{\pi}{2^n+1}$

$Step( III)=$

$P\,(n+1) \, = \, \sqrt{2+P(n)} \, = \, \sqrt{2+2cos\,\dfrac{\pi}{2^{n+1}}}$

$=\, \sqrt{2\left ( 1+cos\frac{\pi}{2^{n+1}} \right )} \, = \, \sqrt{2\, \cdot \, 2cos^2\, \frac{\pi}{2^{n+2}}}$

$= \, 2cos\, \dfrac{\pi}{2^{n+2}}$

Thus

$P \, (n+1)$ holds good.

$\therefore P$ holds for every natural number

$n$ .

Prove the assertions of the following problems
Prove that the expression $n^3 \, - \, n$ is divisible by $24$ for any odd $n$ .

$n = 1$ ----->

$n (n² − 1) = 1 (1 − 1) = 1 (0) = 0$

$n = 3$ ----->

$n (n² − 1) = 3 (9 − 1) = 3 (8) = 24$

We can see that this is true for

$n = 1$ and

$3$

Assume statement is true for

$n = 2k−1$ . Then we must show it is true for

$n = 2k+1$

Statement true for

$n = 2k-1$

$(2k−1) ((2k−1)² − 1) = 24m$ , for some integer m

$(2k−1) ((2k−1)² - 1) = 24m$

$(2k−1) ((2k−1−1)(2k−1+1)) = 24m$

$(2k−1) (2k−2) (2k) = 24m$

$8k³ − 12k² + 4k = 24m$

When

$n = 2k+1:$

$n (n^2 + 1) = (2k+1) ((2k+1)^2 − 1)$

$= (2k+1) ((2k+1−1)(2k+1+1))$

$= (2k+1) (2k) (2k+2)$

$= 8k^3 + 12k^3 + 4k$

$= (8k^3 − 12k^2 + 4k) + 24k^2$

$= 24m + 24k^2$

$= 24 (m + k²) -----> divisible\ by\ 24$

Statement true for

$n = 2k−1$ statement true for

$n = 2k+1$

Therefore, by principles of mathematical induction, statement is true for all odd positive integers n.

Let $u_1$ = 1, $u_2$ = 1 and $u_{n \, + \, 2} \, + \, u_n \, for \, n\geq1$ .
Use mathematical induction to show that
$u_n\, = \, \dfrac{1}{\sqrt{(5)}}\left [ \left ( \dfrac{1 \, + \, \sqrt{5}}{2} \right )^n \, - \, \left ( \dfrac{1 \, - \, \sqrt{5}}{2} \right )^n \right ]$ for all $n \, \geq \, 1$

We have to prove

$u_n \, = \, \dfrac{1}{\sqrt{(5)}}\left [ \left ( \dfrac{1 \, + \, \sqrt{5}}{2} \right )^n \, - \, \left ( \dfrac{1 \, - \, \sqrt{5}}{2} \right )^n \right ]$
for all n

$\geq$ 1
We obviously have

$u_1 \, = \, 1 \, = \, \dfrac{1}{\sqrt{(5)}}\left [\dfrac{1 \, + \, \sqrt{5}}{2} \, - \, \dfrac{1 \, - \, \sqrt{5}}{2} \right ]$
and

$u_2 \, = \, 1 \, = \, \dfrac{1}{\sqrt{(5)}}\left [\left (\dfrac{1 \, + \, \sqrt{5}}{2} \right )^2 \, - \, \left (\dfrac{1 \, - \, \sqrt{5}}{2} \right )^2 \right ]$
Hence (1) holds for n = 1 and n = 2
Now assume

$u_k \, = \, \dfrac{1}{\sqrt{(5)}}\left [\left (\dfrac{1 \, + \, \sqrt{5}}{2} \right )^k \, - \, \left (\dfrac{1 \, - \, \sqrt{5}}{2} \right )^k \right ]$
( k = 1, 2, 3, ....m)
Now

$u_{m \, + \, 2} \, = \, u_{m \, + \, 1} \, + \, u_m$ for m

$\geq$ 1

$\Rightarrow u_{m \, + \, 1} \, = \, u_m \, + \, u_{m \, - \, 1}$ for m

$\geq$ 2
Hence by induction hypothesis on

$u_k$ , we have

$u_{m \, + \, 1} \, = \, u_m \, + \, u_{m \, - \, 1}$

$= \, \dfrac{1}{\sqrt{(5)}}\left [\left (\dfrac{1 \, + \, \sqrt{5}}{2} \right )^m \, - \, \left (\dfrac{1 \, - \, \sqrt{5}}{2} \right )^m \right ] \, + \, \dfrac{1}{\sqrt{(5)}}\left [\left (\dfrac{1 \, + \, \sqrt{5}}{2} \right )^{m \, - \, 1} \, - \, \left (\dfrac{1 \, - \, \sqrt{5}}{2} \right )^{m \, - \, 1} \right ]$

$= \, \dfrac{1}{\sqrt{(5)}} \left [ \left (\dfrac{1 \, + \, \sqrt{5}}{2} \right )^{m \, - \, 1}\left \{\dfrac{1 \, + \, \sqrt{5}}{2} \, + \, 1 \right \} \, - \, \left (\dfrac{1 \, - \, \sqrt{5}}{2} \right )^{m \, - \, 1}\left \{\dfrac{1 \, - \, \sqrt{5}}{2} \, + \, 1 \right \} \right ]$

$= \, \dfrac{1}{\sqrt{(5)}} \left [\left (\dfrac{1 \, + \, \sqrt{5}}{2} \right )^{m \, - \, 1}\left ( \dfrac{6 \, + \, 2\sqrt{5}}{4} \right ) \right ] \, - \, \left (\dfrac{1 \, - \, \sqrt{5}}{2} \right )^{m \, - \, 1}\left ( \dfrac{6 \, - \, 2\sqrt{5}}{4} \right )$

$= \, \dfrac{1}{\sqrt{(5)}} \left [\left (\dfrac{1 \, + \, \sqrt{5}}{2} \right )^{m \, - \, 1}\left ( \dfrac{1 \, + \, \sqrt{5}}{2} \right ) \right ] \, - \, \left [\left (\dfrac{1 \, - \, \sqrt{5}}{2} \right )^{m \, - \, 1}\left ( \dfrac{1 \, - \, \sqrt{5}}{2} \right ) \right ]$

$= \, \dfrac{1}{\sqrt{(5)}} \left [\left (\dfrac{1 \, + \, \sqrt{5}}{2} \right )^{m \, + \, 1}\left ( \dfrac{1 \, - \, \sqrt{5}}{2} \right )^{m \, + \, 1} \right ]$
Thus the formula (1) hold for k = m + 1
Hence(1)holds for all positive integers n by induction

For any natural number n > 1
$\dfrac {1}{n \, + \, 1} \, + \, \dfrac {1}{n \, + \, 2} \, + \, ..... \, + \, \dfrac {1}{2n} > \dfrac {13}{14}$

Let

$S_n \, = \, \dfrac{1}{n+1}\, + \,\dfrac{1}{n+2} \, + \, ....\, + \dfrac{1}{2}$
For

$n \, = \, 2$ , we have

$\dfrac{1}{3}\, + \, \dfrac{1}{4} \, = \, \dfrac{7}{12}\, = \, \dfrac{14}{24}>\dfrac{13}{24}$
Hence the inequality holds for

$n\, = \, 2$
Now assume

$S_m>\dfrac{13}{24}$ for some positive integer

$m>1$
We have

$S_m\, = \, \dfrac{1}{m+1} \, + \,\dfrac{1}{m+2} \, + \, ....\, + \, \dfrac{1}{2m}$
and

$S_{m+1} \, = \, \dfrac{1}{m+2} \, + \, \dfrac{1}{m+3}\, + \, ....\, + \,\dfrac{1}{2m}\, + \, \dfrac{1}{2m+1}\, + \, \dfrac{1}{2m+2}$

$\therefore \, S_{m+1} \, - \, S_m \, = \, \dfrac{1}{2m+1}\, + \dfrac{1}{2m+2}\, - \, \dfrac{1}{m+1}$
that is ,

$\therefore \, S_{m+1} \, - \, S_m \, = \, \dfrac{1}{2(m+1)(2m+1)}>0$ for any natural number m . It follows that

$\therefore \, S_{m+1} \, - \, S_m \, >0,\,\, that\,\, is \,\, S_{m+1}>S_m$
But

$S_m >\dfrac{13}{24}$ by our assumption

$\therefore \, S_{m+1}>\dfrac{13}{24}$
Hence by mathematical induction . The given inequality holds for all natural numbers n >1

Given $a_1\, = \, \dfrac{1}{2}\left ( a_0 \, + \, \dfrac{A}{a_0} \right ) \, , \,a_2\, = \, \dfrac{1}{2}\left ( a_1 \, + \, \dfrac{A}{a_1} \right )$
and $a_{n \, + \, 1}\, = \, \dfrac{1}{2}\left ( a_n \, + \, \dfrac{A}{a_n} \right )$
Find $n\geq \, 2$ where a > 0, A > 0, prove that
$\dfrac{a_n \, - \, \sqrt{A}}{a_n \, + \, \sqrt{(A)}}\, = \, \left (\dfrac{a_1 \, - \, \sqrt{A}}{a_1 \, + \, \sqrt{(A)}} \right )^{2^{n \, - \, 1}}$
using mathematical induction.

We have prove

$\dfrac{a_n \, - \sqrt{A}}{a_n \, + \, \sqrt{(A)}} \, = \, \left (\dfrac{a_1 \, - \sqrt{A}}{a_1 \, + \, \sqrt{(A)}} \right )^{2^n \, - \, 1}$ for all

$n\epsilon N$
Clearly for n = 1, equality (1) holds .
Now from the given relation

$a_2 \, = \, \dfrac{1}{2}\left ( a_1 \, + \, \dfrac{A}{a_1} \right )$
we have

$\dfrac{a_2}{ \sqrt{(A)}} \, = \, \dfrac{1}{2\sqrt{(A)}} \left ( a_1 \, + \, \dfrac{A}{a_1} \right ) \, = \, \dfrac{a_1^2 \, + \, A}{2a_1\sqrt{(A)}}$
Using componendo and dividendo , this gives

$\dfrac{a_2 \, - \, \sqrt{A}}{a_2 \, + \, \sqrt{(A)}} \, = \, \dfrac{a_1^2 \, + \, A \, - \, 2a_1\sqrt{A}}{a_1^2 \, + \, A \, + \, 2a_1\sqrt{(A)}} \, = \, \left (\dfrac{a_1 \, - \, \sqrt{A}}{a_1 \, + \, \sqrt{(A)}} \right )^2$
Hence the equality (1) holds for n = 1,
Now assume that (1) holds for n = m(m

$\geq$ 2), that is, we assume

$\dfrac{a_m \, - \, \sqrt{A}}{a_m \, + \, \sqrt{(A)}} \, = \, \left (\dfrac{a_1 \, - \, \sqrt{A}}{a_1 \, + \, \sqrt{(A)}} \right )^{2^m\, - \, 1}$ ........(2)
Then by the given relation

$a_{n\, + \, 1} \, = \, \dfrac{1}{2} \left ( a_n \, + \, \dfrac{A}{a_n} \right )$ we have

$\dfrac{a_{m \, + \, 1}}{\sqrt{(A)}} \, = \,\dfrac{1}{2\sqrt{(A)}}\left ( a_m \, + \, \dfrac{A}{a_m} \right )$ (m

$\geq$ 2) so that

$\dfrac{a_{m \, + \, 1 \, - \,\sqrt{A}}}{a_{m \, + \, 1}\sqrt{(A)}} \, = \,\dfrac{\dfrac{1}{2}\left ( a_m \, + \, \dfrac{A}{a_m} \right ) \, - \, \sqrt{A}}{\dfrac{1}{2}\left ( a_m \, + \, \dfrac{A}{a_m} \right ) \, + \, \sqrt{A}}$

$= \, \dfrac{am^2 \, - \, 2a_m \sqrt{A} \, + \, A}{am^2 \, + \, 2a_m \sqrt{A} \, + \, A}$

$= \, \left (\dfrac{a_m \, - \, \sqrt{A}}{a_m \, + \, \sqrt{(A)}} \right )^2 \, = \, \left (\dfrac{a_n \, - \, \sqrt{A}}{a_n \, + \, \sqrt{(A)}} \right )^{2^m}$ by (2)
Hence the equality (1) holds for n = m + 1
Hence it holds for all

$n\epsilon N$

Applying the principal of mathematical induction of prove that
$\dfrac{2 \, sin \, \theta }{cos \, \theta \, + \, 3\, cos \, \theta} \, + \, \dfrac{2 \, sin \, \theta }{cos \, \theta \, + \, 5\, cos \, \theta} \, + \, \dfrac{2 \, sin \, \theta }{cos \, \theta \, + \, cos \,(2n \, + \, 1) \theta}$
= tan ( n + 1) $\theta \, - \, tan \theta$

$tan A \, - \, tan B \, = \, \dfrac{sin \, (A \, - \, B)}{cos \, A \, cos \, B}$
for P (1),
L.H.S. =

$\dfrac{2 \, sin \, \theta}{cos \, \theta \, + \, cos \, 3 \theta} \, = \, \dfrac{2 \, sin \, \theta}{2 \, cos \, \theta \, \cdot \, cos \, 2 \theta}$

$= \, \dfrac{sin \, (2 \theta \, - \, \theta)}{cos \, \theta \, cos \, 2 \theta} \, = \, tan \, 2 \theta \, - \, tan \, \theta$
R.H.S. =

$tan \, 2 \theta \, - \, tan \, \theta$ for n = 1
thus P(1) holds good. Assume P(n).
P(n + 1) = P(n) +

$\dfrac{2 \, sin \, \theta}{cos \, \theta \, + \, cos \, (2n \, + \, 3) \, \theta}$

$= \, P(n) \, + \, \dfrac{2 \, sin \, [(n + 2) \, \theta \, - \, (n + 1) \, \theta]}{2 \, cos \, (n + 2) \, \theta \, cos \, (n + 1) \, \theta}$

$= \, [tan \, (n + 1) \, \theta \, - \, tan \, \theta] \, + \, [tan \, (n + 2) \, \theta \, - \, tan \, (n + 1) \, \theta]$

$= \, tan \, (n + 2) \, \theta \, - \, tan \, \theta$
Hence P (n + 1) also holds good

Prove that induction that
sin x + sin 3x + .. + sin(2n - 1) x = $\dfrac {sin^2 nx}{sin x}$

sin x =

$\dfrac{sin^2 \, x}{sin \, x} \, = \, sin \, x \, \ \therefore ,$ p(1) is true
Assume p(m) holds good
p ( m + 1) = p(m) + sin (2m + 1) x
=

$\dfrac{sin^2 mx }{sin x} \, + \, sin\, (2m \, + \, 1)\, x$
=

$\dfrac {1}{2 sin x} [ 2sin^2$ mx + 2 sin x sin (2m +1)x]
=

$\dfrac {1}{2 sin x}$ [(1 - cos 2mx) + cos2mx - cos (2m + 2)x]
=

$\dfrac {1}{2 sin x}$ [1 - cos (2m + 2)x]
=

$\dfrac {1}{2 sin x} \, [2 sin ^2 (m + 1)x] \, = \, \dfrac{sin^2 \, (m \, + \, 1)x}{sinx}$
Thus p(m +1) also holds good

Prove that mathematical induction that
$\dfrac{1}{1 \, + \, x} \, + \, \dfrac{2}{1 \, + \, x^2} \, + \, \dfrac{4}{1 \, + \, x^4} \, + \, ... \, + \, \dfrac{2^n}{1 \, + \, x^{2^n}}$
$= \, \dfrac{1}{1 \, - \, }\dfrac{2^{n \, + \, i}}{1 \, + \, x^{2^{n \, + \, i}}}$

P(0) =

$\dfrac{1}{1 \, + \, x}, \dfrac{1}{1 \, - \, x} \, + \, \dfrac{2}{1 \, + \, x^2} \, = \, \dfrac{1}{1 \, - \, x} \, - \, \dfrac{2}{x^2 \, - \, 1}$

$= \, \dfrac{x \, + \, 1 \, - \, 2}{(x^2 \, - \, 1)} \, = \dfrac{1}{1 \, + \, x}$
Thus P(0) holds goods. Assume P(n)
P(n + 1) = P(n) +

$\dfrac{2^{n \, + \, 1}}{1 \, + \, x^{2^{n \, + \, 1}}}$

$= \, \dfrac{1}{1 \, + \, x} \, + \, \dfrac{2^{n \, + \, 1}}{1 \, - \, x^{2^{n \, + \, 1}}} \, + \, \dfrac{2^{n \, + \, 1}}{1 \, + \, x^{2^{n \, + \, 1}}}$

$= \, \dfrac{1}{1 \, + \, x} \, + \, \dfrac{2^{n \, + \, 1}(1 \, + \, 1)}{1 \, - \, (x^{2^{n \, + \, 1}})^2} \, = \, \dfrac{1}{x \, - \, 1} \, + \, \dfrac{2^{n \, + \, 2}}{1 \, - \, x^{2^{n \, + \, 2}}}$

Observing that $1^3 \, = \, 1, \, 2^3 \, = \, 3 \, + \, 5,$
$3^3$ = 7 + 9 + 11, $4^3$ = 13 + 15 + 17 + 19
Find a general formula for the cube of natural number n and prove it by the principle of mathematical induction

$1^3, \, 2^3, \, .... \, n^3$ have 1,2,..., n terms respectively in their expression and these are in A.P. of d = 2,Also term of these are 1,3,7,13,... Hence first term of

$n^3$ is

$t_n$ of above series, which by method of

$n^3$ is

$t_n$ of above series, which by method of differences is

$n^2$ - n + 1.

$\therefore \, \, \, n^3 \, = \, (n^2 \, - \, n \, + \, 1) \, + \, (n^2 \, - \, n \, + \, 3) .... n \, terms \, \, \, .... (1)$
Clearly P(1),P(2) hold good. Assume P(n) =

$n^3$ .
Replacing by

$(n \, + \, 1)$ in (1).
P (n + 1)

$= \, [(n \, + \, 1)^2 \, - \, (n \, + \, 1) \, + \, 1] \, + \, [(n \, + \, 1)^2 \, - \, (n \, + \, 1) \, + \, 3] .... (n \, + \, 1) \, terms$

$= \, (n \, + \, 1)^2 \, \cdot \, (n \, + \, 1) \, - \, (n \, + \, 1) \, (n \, + \, 1) \, + \, [1 \, + \, 3 \, + \, 5 \, + \, ... \, (n \, + \, 1) \, terms]$

$= \, (n \, + \, 1)^3 \, - \, (n \, + \, 1)^2 \, + \, (n \, + \, 1)^2$
[Sum of an A.P. of (n + 1) terms]

$\therefore \, \, \, P (n \, + \, 1) \, = \, (n \, + \, 1)^3$
For k = 1 we will have two terms for m = 0 and m= 1 in the sigma.
L.H.S. =

$(n \, - \, 0) \, \dfrac{r!}{0!} \, + \, (n \, - \, 1) \, \dfrac{(r \, + \, 1)!}{1!}$

$= \, n \, [(r \, + \, 1)! \, + \, r!] \, - \, (r \, + \, 1)!$

$= \, r! \, [n (r \, + \, 2) \, - \, (r \, + \, 1)]$
and R.H.S. =

$\dfrac{(r \, + \, 2)!}{1!} \, \left[\dfrac{n}{r \, + \, 1} \, - \, \dfrac{1}{r \, + \, 2}\right]$

$= \, (r \, + \, 2)! \, \left[\dfrac{n(r \, + \, 2) \, - \, (r \, + \, 1)}{(r \, + \, 2) \, (r \, + \, 1)}\right]$

$= \, r! \, [n (r \, + \, 2) \, - \, (r \, + \, 1)]$
Hence L.H.S. = R.H.S. for k = 1
Now let the formula hold for k = s, that is,
Let

$\sum_{m \, = \, 0}^{s} \, \dfrac{(n \, - \, m) \, (r \, + \, m)!}{m!}$

$= \, \dfrac{(r \, + \ s \, + \, 1)!}{s!} \left[\dfrac{n}{r \, + \, 1} \, - \, \dfrac{s}{r \, + \, 2}\right]$ ....(1)
Add next term corresponding to m = s + 1
Adding

$\dfrac{(n \, - \, s \, - \, 1)(r \, + \, s \, 1)!}{(s \, + \, 1)!}$ to both sides, we get

$\sum_{m \, = \, 0}^{s \, + \, 1} \, \dfrac{(n \, - \, m) \, (r \, + \, m)!}{m!} \, = \, \dfrac{(r \, + \ s \, + \, 1)!}{s!}$

$\left[\dfrac{n}{r \, + \, 1} \, - \, \dfrac{s}{r \, + \, 2}\right] \, + \, \dfrac{(n \, - \, s \, - \, 1)(r \, + \, s \, 1)!}{(s \, + \, 1)!}$

$= \, \dfrac{(r \, + \ s \, + \, 1)!}{(s \, + \, 1)!} \, \left[\dfrac{(s \, + \, 1)n}{r \, + \, 1} \, - \, \dfrac{s(s \, + \, 1)}{r \, + \, 2} \, + \, n \, - \, s \, - \, 1 \right]$

$= \, \dfrac{(r \, + \ s \, + \, 1)!}{(s \, + \, 1)!} \, \left[ n\left[\dfrac{s \, + \, 1}{r \, + \, 1} \, + \, 1\right] \, - \, (s \, + \, 1) \left[\dfrac{s}{r \, + \, 2} \, + \, 1\right] \right]$

$= \, \dfrac{(r \, + \ s \, + \, 2) \, (r \, + \ s \, + \, 1)!}{(s \, + \, 1)!} \, \left[\dfrac{n}{r \, + \, 1} \, - \, \dfrac{s \, + \, 1}{r \, + \, 2}\right]$

$= \, \dfrac{(r \, + \ s \, + \, 2)!}{(s \, + \, 1)!} \,\left[\dfrac{n}{r \, + \, 1} \, - \, \dfrac{s \, + \, 1}{r \, + \, 2}\right]$
Hence the formula holds for k = s + 1 and so by induction the formula holds for all natural mumbers k.

Explain the method of mathematical induction and use it to show that $11^{n \, + \, 2} \, + \, 12^{2n \, + \, 1}$ where n is natural number is divisible by 133

Let

$A_n \, = \, 11^{n \, + \,2} \, 12^{2n\, + \, 1}$
The assertion is valid for

$n \, = \, 1$ Since,

$A_1 \, = \, \left (11 \right )^{1 \, + \,2} \, + \, \left (12 \right )^{2\, + \, 1} \, = \, 3059 \, = \, 133 \, \times \, 23$
Assume that the assertion holds for n = m ,that is, let

$A_m \, = \, 11 ^{m \, + \,2} \, + \, 12 ^{2m\, + \, 1} \, = \, 133 \, K$ ..............................(1)
where K is a positive integer.
Then

$A_{m \, + \,1} \, = \, 11^{m \, + \, 3} \, + \, 12^{2(m \, +\, 1)\, +\, 1}$

$11^{m \, + \, 3} \, + \, 12^{2m\, +\, 3}$

$11.11^{m \, + \, 2} \, + \, 144.12^{2m\, +\, 1}$

$11 \left ( 133K \, - \, 12^{2m \, + \, 1} \right )\, + \, 144.12^{2m\, +\, 1}$ ............................by (1)

$11.\left ( 133K \, \right )\, + \, \left (144 \, - \, 11 \right ) \,12^{2m\, +\, 1}$

$11.\, 133K \, + \, 133 \,. \, 12^{2m\, +\, 1}$

$133 \, \, \left (11K \,+ \, 12^{2m\, +\, 1} \right )$
This shows that

$A_{m \, + \, 1}$ is divisible by 133. Hence by induction,

$A_n$ is divisible by 133 for all natural number n.

If $x \, + \, y \, = a \, + \, b , x^2 \, + \, y^2 \, = \, a^2 \, + \, b^2$ then prove that the mathematical induction that $x^n \, + \, y^n \, = \, a^n \, + \, b^n$ for all the natural number n .

Clearly p(1) and p(2) hold. Assume p(n).

$p(n + 1) =$

$x^{n +1} \, + \, y^{n +1}$ = x.

$x^n$ + y.

$y^n$

$= x(a^n \, + \, b^n \, - \, y^n) \, + \, y.(a^n \, + \, b^n \, - \, x^n)$

$= (a^n \, + \, b^n)$

$(x + y) - xy$

$(x^{n-1} \, + \, y^{n \, - \, 1})$
Now from given relations

$(x \, + \, y)^2 \, - \, (x^2 \, + \, y^2) \, = \, (a \, + \, b)^2 \, - \, (a^2 \, + \, b^2)$

$\therefore$

$2xy = 2ab\ or\ xy = ab$
Hence from (1)

$p(n + 1) =$

$(a^n \, + \, b^n)$

$(a + b)$ -

$ab(a^{n \, - \, 1} \, + \, b^{n \, - \, 1})$ by p(n - 1)

$= \, a^{n \, + \, 1} \, + \, b^{n \, + \, 1}$
Above shows that

$p(n + 1)$ also holds good.

If $y \, = \, \dfrac{log \, \, x}{x}$ , then prove by mathematical induction
$y_n \, = \, \dfrac{(-1)^n \, (n!)}{x^{n \, + \,1}}\left [ log \, \, x \, - \, 1 \, -\dfrac{1}{2}\, - \, ....-\, \dfrac{1}{n} \right ]$

p (1) =

$y_1 \, = \, \dfrac{1}{x^2}$ (1 - log x)

$= \, \dfrac{(-1)^1 , 1!}{x^{1 \, + \, 1}}$ (log x - 1)
Assume p (n); i.e.,

$y_n \, = \, \dfrac{(-1)^n \, (n!)}{x^{n \, + \, 1}} \left[log \, x \, - \, 1 \, - \, \dfrac{1}{2} \, - \, ... \, - \, \dfrac{1}{n} \right]$
Now p (n + 1) =

$y_{n \, + \, 1} \, = \, \dfrac{d}{dx} (y_n)$

$= \, \dfrac{d}{dx} \, \dfrac{(-1)^2 \, (n!)}{x^{n \, + \, 1}} \left[log \, x \, -1 \, - \, \dfrac{1}{2} \, - \, ... \, - \, \dfrac{1}{n} \right ]$ by
Differentiating as product

$y_{n \, + \, 1} \, = \, (-1)^n \, (n!)$

$\times \, \left[\dfrac{-(n \, + \, 1)}{n^{n \, + \, 2}} \, \left\{log \, x \, - \, 1 \, - \, \dfrac{1}{2} \, - \, ... \, \dfrac{1}{n} \right \}\ \, + \, \dfrac{1}{x^{n \, + \, 1}}.\dfrac{1}{x} \right]$

$= \, \dfrac{(-1)^{n \, + \, 1} \, (n \, + \, 1)!}{x^{n \, + \, 2}} \, \times \, \left[log \, x \, - \, 1 \, - \, \dfrac{1}{2} \, - \, ... \, - \, \dfrac{1}{n} \, - \, \dfrac{1}{n \, + \, 1} \right]$
Thus p (n + 1) also holds good.

If $x^3$ = x + 1, then show that
$x^{3n} \, = \, a_n x \, + \, b_n \, - \, c_n x^{-1}$
Where $a_{n \, + \, 1} \, = \, a_n b_n :b_{n \, + \, 1} \, = \, a_n \, + \, b_n \, + \, c_n$
and $c_{n \, + \, 1} \, = \, a_n \, + \, c_n$

For n = 1, we have

$x^3$ = x + 1 = 1. x + 1 + 0.

$x^{-1}$
so that

$a_1 \, = \, 1, \, b_1 \, = \, 1, \, c_1 \, = \, 0.$
For n = 2,

$x^6 \, = \, (x^3)^2 \, = \, (x \, + \, 1)^2$

$= \, x^2$ + 2x + 1 =

$\dfrac{x^3}{x}$ + 2x + 1

$= \, \dfrac{x \, + \, 1}{x}$ + 2x + 1. = 2x + 2 + 1.

$x^{-1}$
so that

$a_2 \, = \, 2, \, b_2 \, = \, 2, \, c_2 \, = \, 1.$
Here

$a_1 \, + \, b_1$ = 1 + 1 = 2 =

$a_2.$
and

$a_1 \, + \, c_1$ = 1 + 0 =

$c_2.$
Hence the assertion is valid for n = 2.
Let us now assume that

$x^{3m} \, = \, a_m x \, + \, b_m \, + \, c_m x^{-1}$
where

$a_{m \, + \, 1} \, = \, a_m \, + \, b_m; \, b_{m \, + \, 1} \, = \, a_m \, + \, b_m \, + \, c_m.$

$c_{m \, + \, 1} \, = \, a_m \, + \, x_m.$ Then

$x^{3(m \, + \, 1)} \, = \, x^{3m}. \, x^3 \, = \, (a_mx \, + \, b_m \, + \, c_m x^{-1})(x \, + \, 1)$

$= \, a_mx^2 \, + \, b_mx \, + \, c_m \, + \, a_mx \, + \, b_m \, + \, c_mx^{-1}$

$= \, a_m. \, \dfrac{x^3}{x} \, + \, (a_m \, + \, b_m) \, x \, + \, b_m \, + \, c_m \, + \, c_mx^{-1}$

$= \, a_m \, \dfrac{(x \, + \, 1)}{x} \, a_{m \, + \, 1} x \, + \, b_m \, + \, c_m \, + \, c_mx^{-1}$

$= \, a_m \, + \, a_mx^{-1} \, + \, a_{m \, + \, 1} x \, + \, b_m \, + \, c_m \, + \, c_mx^{-1}$

$= \, a_{m \, + \, 1} x \, + \, (a_m \, + \, b_m \, c_m) \, + \, (a_m \, + \, c_m)x^{-1}$

$= \, a_{m \, + \, 1} x \, + \, b_{m \, + \, 1} \, + \, c_{c \, + \, 1}x^{-1}$
Hence assertion holds for n = m + 1
Therefore by mathematical induction, the assertion holds for all natural numbers n.

If $x_1x_2x_3 \, .... \, x_n \ = \, 1(x_1 \, > \, 0, i \, = \, 1,2,.....n),$ prove that $x_1 \, + \, x_2 \, + \, .....x_n \, \geq \, n \, (n \, \geq \, 2)$

$x_1x_2$ = 1, then

$x_1 \, + \, x_2 \, = \, (\sqrt{x_1} \, - \, \sqrt{x_2})^2 \, + \, 2\sqrt{x_1x_2} \, \geq \, 2\sqrt{x_1x_2} \, = \, 2,$
so that the inequality holds true for n = 2.
Now assume that the sum of any m positive numbers whose product is 1 is greater than or equal to m and let

$x_1,x_2,....,x_m, x_{m \, + \, 1}$ be (m + 1) positive integers such that

$x_1.x_2...x_m.x_{m \, + \, 1} \, = \, 1$
We shall prove

$x_1 \, + \, x_2 \, + \, ... \, + \, + \, x_m \, + \, x_{m \, + \, 1} \, = \, 1$
We shall prove

$x_1 \, + \, x_2 \, + \, ... \, + \, x_m \, + \, x_{m \, + \, 1} \, \geq \, m \, + \, 1.$
If each

$x_i$ (i = 1,2,...,m + 1) is 1, then

$x_1 \, + \, x_2 \, + \, ... x_m \, + \, x_{m\, + \, 1} \, = \, m \, + \,,$ so that in this case the inequality holds true.
If

$x_i's$ are not all 1, then among them there will be a number greater than 1 and a number less than 1. Let

$x_m$ > 1 and

$x_{m \, + \, 1}$ < 1.
We have

$x_1x_2...x_{m \, - \, 1}(x_mx_{m \, + \, 1})$ = 1.
This is a product of m numbers and so by our induction hypothesis, we can say that

$x_1 \, + \, x_2 \, + \, ... \, x_{m \, - \, 1} \, x_mx_{m \, + \, 1} \, \leq \, m$
But then

$x_1 \, + \, x_2 \, + \, ... \, x_{m \, - \, 1} \, + \, x_m \, + \, x_{m \, + \, 1}$

$\leq m \, - \, x_mx_{m \, + \, 1} \, + \, x_m \, + \, x_{m \, + \, 1}$ by

$= \, m \, + \, 1 \, +(x_m \, - \, 1)(1 \, - \, x_{m \, + \, 1}) > \, m \, + \, 1$
Since

$x_m$ -1>0 and 1 -

$x_{m \, + \, 1} > 0$ by
so that

$(x_m \, - \, 1)(1 \, - \, x_{m\, + \,1 }) > \, 0.$
This completes the proof.

Prove that the mathematical induction that
$1 \, + \, \dfrac{1}{2} \, + \, \dfrac{1}{3} \, + \, \dfrac{1}{4} + ...\ , \dfrac {1}{2^n} \geq 1 \, + \, \dfrac {n}{2}$
for each no negative interger n

p(0) , p(1) , clearly holds goods

$p(2) \, = \, 1 \, + \, \dfrac{1}{2}\, + \, \dfrac{1}{3} \, + \, \dfrac{1}{4 }\, = \, \dfrac{25}{12}\, > \dfrac{24}{12} \, = \, 2 \, 1 \, + \, \dfrac{2}{2}$
Thus p(2) holds goods ,
p(n + 1) =

$\left ( 1 \, + \, \dfrac{1}{2} \, + \, \dfrac{1}{3} \, + \, \dfrac{1}{4} \, ...... + \, \dfrac{1}{2^n }\right) \, + \dfrac{1}{2^{n + 1}}$
=p(n) +

$\left ( \dfrac{1}{1 \, + 2^n} \, + \, \dfrac{1}{2 \, + 2^n}\, + \, \dfrac{1}{2^n \, + 2^n} \right)$
The 2 nd bracket contain 1 , 2 , 3 , .........

$2^n$ . i.e. 2 terms Each term of this bracket

$> \dfrac{1}{2^n \, + 2^n} \, + \, \dfrac{1}{2 \, \cdot 2^n} \ \dfrac{1}{2^{n =1}}$

$\therefore p(n +1) \geq \, \left ( 1 \, + \, \dfrac{n}{2} \right) \, + \, 2^n \, \cdot \, \dfrac{1}{2^{n+1}}$

$= \, 1 \, + \, \dfrac{n }{ 2}\, + \, \dfrac {1}{2} \, = \, 1\, + \, \dfrac{n + 1 }{2}$

$\therefore \, \, p ( n \, + \, 1) \, \geq \, 1\, + \, \dfrac{n + 1 }{2}$
Thus p(n ) is universally true

Prove $^{2n} C _n \, > \, \dfrac{4^n }{n + 1}$

p(2) holds goods as

$^4C_2 \, = \, 6 \, > \, \dfrac {16}{3}$
Assume p (n ) holda
p(n + 1) =

$^{2n + 2}C _{n +1} \, = \, \dfrac {(2n \, + \, 2)!}{(n \, + \, 1)! (n \, + \, 2)! }$

$= \, \dfrac{(2n \, + \, 2)(2n \, + \, 1) } {(n \, + \, 1) \cdot \, (n \, + \, 2)} \, \cdot \, \dfrac{(2n)!}{n!n!}$
or

$\dfrac {2(2n \, + \, 1)}{(n \, + \, 1)} \, ^{2n }C _n \, \dfrac {2(2n \, + \, 1)}{(n \, + \, 1)}\dfrac {4^n }{n \, + \, 1}$
Now p (n +1) will holds good if we show that

$\dfrac {2(2n \, + \, 1)}{(n \, + \, 1)(n \, + \, 2)} \, \cdot \, 4^n \, > \, \dfrac {4^{n +1}}{(n \, + \, 2)}$
or

$(2n + 1)(n + 2) > 2$

$(n \, + \, 1)^2$
or

$2n ^2 , + \, 5n \, + \, 2 \, > \, 2n^2 \, + \, 4n \, + \, 2 \,$
or 5 > 4 which is true
Hence p(n + 1) holds goods

Prove $\dfrac {2n!}{(n!)^2} \, > \, \dfrac {4n}{2n \, + \, 1}$

we can verify for n = 2 as in part (i)
Assume p(n) i. e.

$\dfrac{(2n)!}{(n!)^2}> \dfrac{4n}{2n \, + \, 1}$
p(n + 1) =

$\dfrac {(2n + \, 2)!}{[(n \, + \, 1)!]^2}$
=

$\dfrac{(2n \, + \, 2)(2n \, + \, 1)}{(n \, + \, 1)^2}\, \dfrac{(2n)!}{(n!)^2}$

$= \, \dfrac{2 (2n \, + \, 1)}{n \, + \, 1} \, \, p (n ) > \dfrac{2 (2n \, + \, 1)}{n \, + \, 1} \, \cdot \, \dfrac{4n }{2n \, + \, 1}$

$= \dfrac{8n }{n \, + \, 1}$
We have prove p( n +1) >

$\dfrac{4 ( n \, + \, 1)}{2n \, + \, 3 }$
or

$\dfrac{8n }{n \, + \, 1} > \dfrac{4 (n \, + \, 1)}{2n \, + \, 1}$
or

$2n (2n +3) -$

$(n^2$

$+ 2n + 1)> 0$
or

$3n^2$

$+ 3n + n - 1 > 0$
or 3

$n (n + 1) + (n - 1) > 0$
Above is true for n > 1

Prove by induction $\forall \, n \, \epsilon \, N \, \, if y \, = \, \dfrac{x}{x^2 \, + \, a^2}$
then $y_n \, = \, \dfrac{(-1)^n \, n!}{a^{n \, + \,1}}sin^{n \, + \, 1} \, \, \theta \, \, cos(n \, + \, 1) \, \, \theta$
where $\theta \, = \, cot^{-1} \dfrac{x}{a}$

x = a cot

$\theta$

$\therefore \, \dfrac{dx}{d\theta} \, = \, -a \, cosec^2 \, \theta$

$y \, = \, \dfrac{x}{x^2 \, + \, a^2} \, = \, \dfrac{a \, cot \, \theta}{a^2 \, cosec^2 \, \theta} \, = \, \dfrac{1}{a} \, sin \, \theta \, cos \, \theta$

$\therefore \, y_1 \, = \, \dfrac{dy}{dx} \, = \, \dfrac{dy}{d\theta}.\dfrac{d\theta}{dx} \, = \, \dfrac{1}{a} \, (cos^2 \, \theta \, - \, sin^2 \, \theta)$

$\left(-\dfrac{1}{a \, cosec^2 \, \theta} \right)$

$= \, -\dfrac{1}{a^2} sin^2 \, \theta \, cos \, 2\theta$
Thus p (n) holds for n = 1.
Assume p (n) holds good.

$\therefore \, y_n \, = \, \dfrac{(-1)^n \, n!}{a^{n \, + \, 1}} \, sin^{n \, + \, 1} \, \theta \, cos \, (n \, + \, 1) \, \theta$
Differentiate both sides w.e.t. x.

$\therefore \, y_{n \, + \, 1} \, = \, \dfrac{(-1)^n \, n!}{a^{n \, + \, 1}}$
[(n +1)

$sin^n \, \theta \, cos \, \theta \, cos(n \, + \, 1)\theta\, -sin^{n \, + \, 1} \, \theta. \, (n \, + \, 1) \, sin \, (n \, + 1) \theta] \, ]\dfrac{d\theta}{dx}$

$= \, \dfrac{(-1)^n \, n!(n \, + \, 1)}{a^{n \, + \, 1}} \, sin^n \, \theta \, cos$ ( n + 1 + 1 )

$\theta \, \left(-1\dfrac{1}{a \, cosec^2 \, \theta} \right)$

$= \, \dfrac{(-1)^n \, (n \, + \, 1)!}{a^{n \, + \, 2}} \, sin^{n \, + \, 2} \, \theta \, cos \, (n \, + \, 2) \, \theta$
Hence p (n)

$\Rightarrow$ p (n + 1).

If 0 < $\alpha < \, \dfrac{\pi}{4(n \, - \, 1)}$ where n > 1 , then prove that $\tan n$ $\alpha \, > \, n \, \tan \, \alpha$

Since

$n > 1 , p(2)$ holds as

$\tan$ 2

$\alpha$ >

$2 \tan$

$\alpha$
when 0 <

$\alpha$

$< \dfrac{\pi}{4}$

$(n = 2)$

$\therefore \, \, \dfrac{2 \tan \, \alpha}{1 \, - \, \tan^2 \, \alpha}$ 2 tan

$\alpha$ as tan

$\alpha$ is +ive and < 1
Now assume P (m) i.e.,

$\tan$

$\alpha$ >

$m \tan$

$\alpha$
Now

$\tan(m + 1)$

$\alpha$ =

$\dfrac{\tan \, m\alpha \, + \, \tan \, \alpha}{1 \, - \, \tan \, m\alpha \,\, \tan \, \alpha}$ ...(1)
Where

$0 \, < \, \alpha \, < \, \dfrac{\pi}{4(m \, + \, 1 \, - \, 1)}$
or

$0 \, < \, \alpha \, < \, \dfrac{\pi}{4m} \, \, or \, \, 0 \, < \, m\alpha \, < \, \dfrac{\pi}{4}$

$\therefore \, \, \tan \, m\alpha \, < \, 1$ and is

$+ive$ . Also

$\tan$

$\alpha$ is < 1 and +ive .
Hence from (1)

$\tan (m + 1)$

$\alpha$ >

$m \tan$

$\alpha$

$+ \tan$

$\alpha$

$= ( m+ 1) \tan$

$\alpha$

$\therefore$

$P(m + 1)$ also holds good .
Hence universally true.

Show by the mathematical induction that
$\dfrac{1}{sin 2 x } \, + \, \dfrac{1}{sin 4 x } \, + \, \dfrac{1}{sin 2^n x } \, = \, cot x \, - \, cot 2^n \, x$

cot A - cot B =

$\dfrac {sin(B - A)}{sin A sin B }$ For p (1) L. H. S. =

$\dfrac{1}{sin 2 x}$

R. H. S. = cot x - cot 2x =

$\dfrac{sin (2x \, - \, x)}{sin x \, sin 2x }\, = \, \dfrac{1}{sin 2x }$

Thus p(1) is holds goods assume p(n) is true for all

p(n + 1) = p(n) +

$\dfrac{1}{sin 2^{n +1}x}$

= p(n) +

$\dfrac {sin 2 ^n x }{sin 2^n x sin{2} ^{n +1}x}$ = p(n) +

$\dfrac {sin(2^{n +1} \, - \, 2^n)x}{sin 2^n \, x \, sin 2^{n + 1}x}$

$(cot \, x \, - \, cot 2^n x ) \, + \, (cot \, 2^n \, - \, cot2^{n +1} x)$

$= cot x \, - \, cot \, 2^{n + 1} x$

Thus the p(n + 1) alos goods holds .

Prove that ${ 1 }^{ 2 }+{ 2 }^{ 2 }+...+{ n }^{ 2 }>\cfrac { { n }^{ 3 } }{ 3 } ,n\epsilon N$

$P(n):{ 1 }^{ 2 }+{ 2 }^{ 2 }+...+{ n }^{ 2 }>\cfrac { { n }^{ k } }{ 3 }$

$P(1):1>\cfrac { { 1 }^{ k } }{ 3 }$

Clearly for

$k=1,1>\cfrac { 1 }{ 3 }$

For

$k=2,P(2):1+{ 2 }^{ 2 }>\cfrac { { 2 }^{ k } }{ 3 }$

$\Rightarrow 5>\cfrac { { 2 }^{ k } }{ 3 }$

For

$k=1,P(2):5>\cfrac { 2 }{ 3 }$ for

$k=1,P(3):1+4+9>\cfrac { 3 }{ 3 }$

For

$k=2,P(2):5>\cfrac { 4 }{ 3 }$ for

$k=2,P(3):14>\cfrac { { 3 }^{ 2 } }{ 3 }$

For

$k=3,P(2):5>\cfrac { 8 }{ 3 }$ for

$k=3,P(3):14>\cfrac { { 3 }^{ 3 } }{ 3 }$

For

$k=4,P(2):5\ngtr \cfrac { 16 }{ 3 }$ for

$k=4,P(3):14\ngtr \cfrac { { 3 }^{ 4 } }{ 3 }$

$P(2)$ is satisfied for

$k=1,2,3$

$\therefore k=3$ , where

$k\epsilon { Z }^{ + }$

Now we want to verify the value for

$k\forall$ positive integral values of n using principle of mathematical induction

Verification:Let

$P(m)$ be true for

$n=m$ then

Prove the following by using the first principle
$1+3 + 3^2 + ......+ 3^{n-1} =\frac{(3^n -1)}{2}$ .

$\\(a)\>check\>p(1)LHS=1\\RHS\>=(\frac{3^1-1}{2})=(\frac{2}{2})=1\\LHS=RHS\\(b)let\>P(K)be\>true\>,then\\1+3+3^2----------+3^{k-1}=(\frac{3^k-1}{2})\\\>now\>for\>p(k+1)\\\>LHS=1+3+3^2-------+3^{K-1}+3^k\\a\>Gp\>with\>(k+1)\>terms\>=(\frac{1\cdot\>(3^{k+1}-1)}{3-1})=(\frac{3^{k+1}}{2})\\\>RHS=(\frac{3^{k+1}}{2})\\\therefore\>LHS\>=RHS\\Hence\>p(k+1)is\>true\>,so\>p(n)is\>True\>for\>all\>n\>\in\>N\>using\>PMI$

Prove the following by PMI for all $n$ belong to $N$
$1\times3+3\times5+5\times7 .........+(2n-1)(2n+1)=\dfrac{n(4n^2+6n-1)}{3}$

When

$n=1$ we have the end term of the series as

$(2*1 -1)(2*1 +1) = 1*3 = 3$
Putting

$n=1$ in the R.H.S of the given equation we have

$\dfrac{1(4*1^2 + 6*1 - 1)}{3} = \dfrac {1(4 + 6 -1)}{3} = 3$
Therefore the equation is valid for

$n=1$
Let the expression be valid for any value

$n=k$ where 'k' belongs to N.
So

$1.3 + 3.5 +.....+(2k-1)(2k+1)=\dfrac{k(4k^2+6k-1)}{3}$ holds true. ---------eqn(1)
Now we have to prove that the equation is valid for n=k+1.
i.e.

$1.3+3.5+....+[2(k+1)-1][2(k+1)+1] = \dfrac{(k+1)(4(k+1)^2+6(k+1)-1)}{3}$ ----eqn(2)
Now L.H.S of equation 2 can be written as

$1.3 + 3.5 +....+(2k-1)(2k+1) + [2(k+1)-1][2(k+1)+1]$ -----------expression(1)
Putting the value of the R.H.S of equation 1 in expression 1 we have

$\dfrac{k(4k^2+6k-1)}{3} + (2k+2-1)(2k+2+1)$

$\dfrac{(4k^3+6k^2-k)}{3} + \dfrac{3(2k+1)(2k+3)}{3}$

$\dfrac{(4k^3+6k^2-k)}{3} + \dfrac{3(4k^2+6k+2k+3)}{3}$

$\dfrac{(4k^3+6k^2-k)}{3}+ \dfrac{(12k^2+24k+9)}{3}$

$\dfrac{(4k^3+6k^2-k+12k^2+24k+9)}{3}$

$\dfrac{(4k^3+18k^2+23k+9)}{3}$

$\dfrac{(4k^3+4k^2+14k^2+14k+9k+9)}{3}$

$\dfrac{{4k^2(k+1)+14k(k+1)+9(k+1)}}{3}$

$\dfrac{(k+1)(4k^2+14k+9)}{3}$

$\dfrac{(k+1)(4k^2+8k+6k+4+6-1)}{3}$

$\dfrac{(k+1){4k^2+8k+4+6k+6-1}}{3}$

$\dfrac{(k+1)(4(k^2+2k+1)+6(k+1)-1)}{3}$

$\dfrac{(k+1){4(k+1)^2+6(k+1)-1}}{3}$ -----expression(2)

But expression 2 is nothing but the R.H.S. of the equation 2.
Therefore by mathematical induction we have proved that the said equation holds true for every value of 'n' where 'n' belongs to N.

Prove that for any positive interger $n, n^3 -n$ is divisible by $6$ .

Let

$f(n)=n^3-n$ .

Now

$f(1)=0$ , divisible by

$6$ .

Again

$f(2)=6$ , divisible by

$6$ .

So the given statement is true for

$n=1,2$ .

Let us take the statement is true for some integer

$k$ .

$\therefore f(k)=k^3-k$ is divisible by

$6$ .

Let

$f(k)=6k_1$ [

$k_1$ being an integer].......(1).

Now

$f(k+1)$

$=(k+1)^3-(k+1)$

$=k^3+3k^2+2k$

$=k^3-k+3k^2=3k$

$=(k^3-k)+3k(k+1)$ .........(2).

$k(k+1)$ is the product of two consecutive integers, which is always divisible by

$2$ .

Then we can write

$k(k+1)=2k_2$ [Where

$k_2$ is some integer.].

Using this and (1) form (2) we get,

$f(k+1)=6k_1+6k_2=6(k_1+k_2)$

$k_1,k_2$ are integers then

$k_1+k_2$ is also an integer.

$f(k+1)$ is also divisible by

$6$ .

So the statement is true for

$n=k+1$ if we assume it to be true for

$n=k$ .

By the principle of mathematical induction the statement is true

$\forall n\in \mathbb{N}$ .

Let p $\geq$ 3 be an integer and $\alpha$ , $\beta$ be the roots of $x^2$ - (p + 1)x + 1 = 0.Using mathematical induction show that $\alpha^n \, + \, \beta^n$
is not divisible by p.

Given

$p \, \geq \, 3$ is an integer and

$\alpha \, , \, \beta$ are the roots of

$x^2$ + (p + 1) x + 1) = 0
so that

$\alpha \, + \, \beta \, = \, p \, + \, 1\, , \, \alpha\beta \, = \, 1$ .....(1)

$\alpha^2 \, + \, \beta^2 \, = \, (\alpha \, + \, \beta)^2\, - \, 2\alpha\beta$
= (p + 1)

$^2$ - 2 = p

$^2$ + 2p - 1 .........(2)
Now p is an integer greater than or equal to 3, therefore from (1) and (2) we conclude that P(1), P(2) hold good and assume that P(n) also holds good i.e., they are all integers .
Now multiplying both sides of
x

$^2$ - (p + 1) x + 1 = 0 by x

$^{n \, - \, 1}$ , we get
x

$^{a \, + \, 1}$ = (p + 1) x

$^n$ + x

$^{n \, - \, 1}$
Its roots are

$\alpha$ and

$\beta$ also

$\alpha^{n \, + \, 1} \, + \, \beta^{n \, + \, 1} \, = \, (p \, + \, 1)(\alpha^n \, + \, \beta^n)\, + \, (\alpha^{n \, - \, 1} \, + \, \beta^{n \, - \, 1})$
Or P(n + 1) = (p + 1) P(n) + P(n - 1) .......(3)
Since P(n) and P(n - 1) hold good, therefore P(n + 1) also holds good
i.e.,

$\alpha^{n \, + \, 1} \, + \, \beta^{n \, + \, 1}$ is an integer
Again

$\dfrac{\alpha \, + \, \beta}{p} \, = \, \dfrac{p \, + \,1}{p} \, = \, 1 \, + \, \dfrac{1}{p}$

$\dfrac{\alpha^2 \, + \, \beta^2}{p} \, = \, \dfrac{p^2 \, + \,2p \, - \, 1}{p} \, = \,(p \, + \, 2 ) \, - \, \dfrac{1}{p}$
From above we conclude that

$\alpha \, + \, \beta$ and

$\alpha^2 \, + \, \beta^2$ are not divisible by p since 1/p is a proper fraction , therefore P (1), P (2) hold good and assume P (n) also holds good i.e.,P (n) is not divisible by p.

$\therefore$ P(n + 1) = (p + 1) P (n) + P (n - 1) from(3)
since each of P (n) and P (n - 1) are not divisible by 3

$\therefore$ P(n + 1) = (p + 1) P (n) + P (n - 1) is also not divisible (3) .
This proves the 2nd part

Prove by induction method that for all n $\geq$ 1
$\int x^ne^x\, dx \, = \, n\, !\, e^x\left [ \dfrac{x^n}{n!} \, - \, \dfrac{x^{n \, - \, 1}}{(n \, - \, 1)!} \, + \, \dfrac{n^{n \, - \, 2!}}{(n \, - \, 2)!} \, + \, .....\, + \, (-1)^n \right ]$

p(1) =

$\int xe^x$ dx =

$x.e^x \, - \int e^2$ . 1dx =

$e^x$ (x - 1)
R.H.S. =

$e^x \, \left(\dfrac{x}{1!} \, - \, \dfrac{1}{0!} \right) \, = \, e^x$ (x - 1)
Thus p(1) holds good. Assume p(n)

$\therefore$ p(n + 1) =

$\int x^{n \, + \, 1} \, e^x$ dx
=

$x^{n+1} \, .e^x$ -(n + 1)

$\int x^n \, e^x$ dx
=

$x^{n \, + \, 1} \, .e^x$ - (n - 1).n!

$e^x \, \left[\dfrac{x^n}{n!} \, - \, \dfrac{x^{n \, - \, 1}}{(n \, - \, 1)!} \, + \, \dfrac{x^{n \, - \, 2}}{(n \, - \, 1)!} \, + \, ... \right]$
= (n + 1)!

$e^x \, \left[\dfrac{x^{n \, - \, 1}}{(n \, + \, 1)!} \, - \, \dfrac{x^n}{n!} \, + \, \dfrac{x^{n \, - \, 2}}{(n \, - \, 1)!} \, - \, ... \right]$
Thus p(n + 1) also holds good.

Let p $\geq$ 3 be an integer and $\alpha$ , $\beta$ be the roots of $x^2$ - (p + 1)x + 1 = 0.Using mathematical induction show that $\alpha^n \, + \, \beta^n$
is an integer

Given

$p \, \geq \, 3$ is an integer and

$\alpha \, , \, \beta$ are the roots of

$x^2$ + (p + 1) x + 1) = 0
so that

$\alpha \, + \, \beta \, = \, p \, + \, 1\, , \, \alpha\beta \, = \, 1$ .....(1)

$\alpha^2 \, + \, \beta^2 \, = \, (\alpha \, + \, \beta)^2\, - \, 2\alpha\beta$
= (p + 1)

$^2$ - 2 = p

$^2$ - (p + 1) x + 1 = 0 by x

$^{n \, - \, 1}$ , we get
x

$^{a \, + \, 1}$ = (p + 1) x

$^n$ + x

$^{n \, - \, 1}$
Its roots are

$\alpha$ and

$\beta$ also

$\alpha^{n \, + \, 1} \, + \, \beta^{n \, + \, 1}$ is an integer
Again

$\dfrac{\alpha \, + \, \beta}{p} \, = \, \dfrac{p \, + \,1}{p} \, = \, 1 \, + \, \dfrac{1}{p}$

$\dfrac{\alpha^2 \, + \, \beta^2}{p} \, = \, \dfrac{p^2 \, + \,2p \, - \, 1}{p} \, = \,(p \, + \, 2 ) \, - \, \dfrac{1}{p}$
From above we conclude that

$\alpha \, + \, \beta$ and

$\alpha^2 \, + \, \beta^2$ are not divisible by p since 1/p is a proper fraction , therefore P (1), P (2) hold good and assume P (n) also holds good i.e.,P (n) is not divisible by p.

$\therefore$ P(n + 1) = (p + 1) P (n) + P (n - 1) from(3)
since each of P (n) and P (n - 1) are not divisible by 3

$\therefore$ P(n + 1) = (p + 1) P (n) + P (n - 1) is also not divisible (3) .
This proves the 2nd part

Let $I_m \, = \, \int_{0}^{\pi}\dfrac{1 \, - \, cos \, \, mx}{1 \, - \, cos \, \, x}dx$
Use mathematical induction to prove that
$I_m \, = \, m\pi \, , \, m0$ , , 1, 2.....

$p(1) \, = \, \int_{0}^{\pi} \, \dfrac{1 \, - \, cos \, x}{1 \, - \, cos \, x}dx=[x]_0^\pi = 1. \pi = \pi$

$p(2) \, = \, \int_{0}^{\pi} \, \dfrac{1 \, - \, cos \, 2x}{1 \, - \, cos \, x}dx \, = \, \int_{0}^{\pi} \, \dfrac{2 \, sin^2 \, x}{2 \, sin^2 \, (x/2)}$

$\int_{0}^{\pi} \, 4.cos^2 \, \dfrac{x}{2}dx$

$= \, \int_{0}^{\pi}2(1 \, + \, cos \, x)dx = 2. \, \pi \, + \, 0 \, = \, 2\pi$
Let us assume that p (m) = mn and we will establish that p(m + 1) = (m + 1)

$\pi$

$\therefore \, p(m \, + \, 1) \, = \,\int_{0}^{\pi} \dfrac{1 \, - \, cos \, (m \, + \, 1)}{1 \, - \, cos \, x}dx \, = \, (m \, + \, 1)\pi$
Now cos (m + 1) x + cos (m - 1) x = 2 cos mx cos x

$\therefore$ -cos(m + 1)x
= cos(m - 1) x - 2 cos mx cos x
or 1 - cos (m + 1)x
= 1 + cos(m - 1) x - 2 cos mx cos x
= -[1 - cos(m - 1) x] + 2 cos mx (1 - cos x) + 2 - 2 cos mx
Now divide throughout by 1 - cos x and integrate

$\therefore$ p(m + 1) (by 1)
= -p(m - 1) + 2 p (m) +

$\int_{0}^{\pi}$ 2 cos mx dx
= -(m - 1)

$\pi$ + 2

$m\pi$ + 2

$[sin \, mx]_0^\pi$
= (m + 1)

$\pi$ + 0 = (m + 1)

$\pi$

By PMC, prove that inequality $n<2^n$ for all $n\in N$

Step 1: prove for

$n=1$

$1 < 2$

Step 2:

$n+1<2(2^n)$

$n<2(2^n)−1$

$n<2^n+2^n−1$

The function

$2^n+2^n−1$ is surely higher than

$2^n−1$ so if

$n<2^n$ is true (induction step),

$n<2^n+2^n−1$ has to be true as well.

Use mathematical induction to prove
$1 + 2 + 3 + \ldots + n < \frac{1}{8}{\left( {2n + 1} \right)^2}$

$1 + 2 + 3 + \ldots + n < \frac{1}{8}{\left( {2n + 1} \right)^2}$
Let

$n = 1$

$\begin{array}{c}P\left( 1 \right):1 < \frac{1}{8}{\left( {2\left( 1 \right) + 1} \right)^2}\\1 < \frac{1}{8}\left( 9 \right)\end{array}$
is true

$\begin{array}{l}n = k\\P\left( k \right):1 + 2 + 3 + \cdots + k < \frac{1}{8}{\left( {2k + 1} \right)^2}\end{array}$
when

$n = k + 1$

$\begin{array}{l}1 + 2 + 3 + \cdots + k + \left( {k + 1} \right) < \frac{1}{8}{\left( {2k + 1} \right)^2} + \left( {k + 1} \right)\\ < \frac{1}{8}\left[ {4{k^2} + 12k + 9} \right]\\ < \frac{1}{8}{\left[ {2k + 3} \right]^2}\\ < \frac{1}{8}{\left[ {2\left( {k + 1} \right) + 1} \right]^2}\end{array}$
Hence proved

If $c>a$ ; show that $cba-abc=99(c-a)$ .

1310506_1037276_ans_ab47e1d83ca6466fa775f8e16302dcc7.jpg

if $a>c$ ; show that $abc-cba=99(a-c)$ .

1310510_1037272_ans_92926905acba4c7a82417c8a67fd8e99.jpg

Using the Mathematical induction, show that for any number $n \geq 2,$
$\dfrac{1}{1 + 2} + \dfrac{1}{1 + 2 + 3} + \dfrac{1}{1 + 2 + 3 + 4} + ..... +\dfrac{1}{1 + 2 + 3 + ... + n} = \dfrac{n - 1}{n + 1}$

$\begin{array}{l}\dfrac{1}{{1 + 2}} + \dfrac{1}{{1 + 2 + 3}} + \dfrac{1}{{1 + 2 + 3 + 4}} + .... + \dfrac{1}{{1 + 2 + 3 + .... + n}} = \dfrac{{n - 1}}{{n + 1}}\\\dfrac{1}{{1 + 2}} + \dfrac{1}{{1 + 2 + 3}} + \dfrac{1}{{1 + 2 + 3 + 4}} + ....\\{T_n} = \dfrac{1}{{1 + 2 + 3 + .....\left( {n + 1} \right)}} = \dfrac{1}{{\dfrac{{\left( {n + 1} \right)\left( {n + 2} \right)}}{2}}}\\ = \dfrac{2}{{\left( {n + 1} \right)\left( {n + 2} \right)}} = 2\left[ {\dfrac{1}{{\left( {n + 1} \right)}} - \dfrac{1}{{\left( {n + 2} \right)}}} \right]\\{T_1} = 2\left[ {\dfrac{1}{2} - \dfrac{1}{3}} \right],{T_2} = 2\left[ {\dfrac{1}{3} - \dfrac{1}{4}} \right],....{T_n} = 2\left[ {\dfrac{1}{{\left( {n + 1} \right)}} - \dfrac{1}{{\left( {n + 2} \right)}}} \right]\\{S_n} = 2\left[ {\dfrac{1}{2} - \dfrac{1}{{\left( {n + 1} \right)}}} \right] = \dfrac{{n - 1}}{{n + 1}}\end{array}$

Let by mathematical induction, for any natural numbers n,
$\dfrac{1}{2.5} + \dfrac{1}{5.8} + \dfrac{1}{8.11} + ..... + \dfrac{1}{(3n - 1)(3n + 2)} = \dfrac{n}{an + b}$ . Find $a+b$

Solution:

$\displaystyle \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + -- + \frac{1}{(3n-1)(3n+2)} = \frac{n}{(an+b)}$

Let

$\displaystyle p(n) = \frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+ ---+\frac{1}{(3n-1)(3n+2)}=\frac{n}{an+b}$

Consider

$a=6, b=4$ form

$= 1$

$\therefore LHS = \dfrac{1}{2.5} = \dfrac{1}{10}$

$RHS = \dfrac{1}{[6(1)+4]} = \dfrac{1}{10}$

Hence, LHS = RHS

$\therefore a+b = 6+4=10$

Prove that ${2.7^n} + {3.5^n} - 5$ is divisible by $24$ for all $n \in N$ .

Let

$P(n): 2.7^n+3.5^n-5$ is divisible by

$24$

We note that

$P(n)$ is true when

$n=1$ , since

$2.7+3.5-5=24$ . which is divisible by

$24$ .

Assume that

$P(k)$ is true.

i.e.

$2.7^k+3.5^k-5=24q$ when

$q\in N$ -------------- ( 1 )

Now, we have to prove that

$P(k+1)$ is true whenever

$P(k)$ is true.

We have

$2.7^{k+1}+3.5^{k+1}-5$

$\Rightarrow$

$2.7^k.7^1+3.5^k.5^1-5$

$\Rightarrow$

$7[2.7^k+3.5^k-5-3.5^k+5]+3.5^k.5-5$

$\Rightarrow$

$7[24q-3.5^k+5]+15.5^k-5$

$\Rightarrow$

$2\times 24q-21.5^k+35+15.5^k-5$

$\Rightarrow$

$7\times 24q-6.5^k+30$

$\Rightarrow$

$7\times 24q-6(5^k-5)$

$\Rightarrow$

$7\times 24q-6(4p)$ [

$(5^k-5)$ is multiple of

$4$ ]

$\Rightarrow$

$7\times 24q-24p$

$\Rightarrow$

$24(7p-q)$

$\Rightarrow$

$24\times r;\,r=7p-q.$ is some natural number ---------- ( 2 )

The expression on the

$R.H.S$ oof ( 1 ) is divisible by

$24$ . Thus

$P(k+1)$ is true whenever

$P(k)$ is true.

Hence, by principle of mathematical induction ,

$P(n)$ is true for all

$n\in N.$

Prove by the principle of mathematical induction that $2^n > n$ for all $n \in N.$

Let

$P(n)$ be the statement:

$2^{n} >n$

$P(1)$ means

$2^{1}>1$ i.e.

$2>1,$ which is true

$\Rightarrow P(1)$ is true.

Let

$P(m)$ be true

$\Rightarrow 2^{m} >m$

$\Rightarrow 2.2^{m} >2.m \Rightarrow 2^{m+1}>2m\geq m+1$

$\Rightarrow 2^{m+1}>m+1$

$\Rightarrow P(m+1)$ is true.

$\therefore$

$2^{n}>n$ for all

$n\in N$

Prove the following by mathematical induction $1 + 2 + 3 + \cdots + n < \frac{1}{8}{\left( {2n + 1} \right)^2}$

$1 + 2 + 3 + \cdots + n < \frac{1}{8}{\left( {2n + 1} \right)^2}$

Let,

$n = 1$

$p(1):\,1 < \frac{1}{8}{\left( {2(1) + 1} \right)^2}$

$1 < \frac{1}{8}\,\,(9)$

$n = k$

$p(k):\,1 + 2 + 3 + ........ + k < \frac{1}{8}{(2k + 1)^2}$

when

$n = k + 1$

$\begin{array}{l}1 + 2 + 3 + \cdots + k + (k + 1)\\ < \frac{1}{8}{(2k + 1)^2} + (k + 1)\\ < \frac{1}{8}\left[ {4{k^2} + 1 + 4k + 8k + 8} \right]\\\, < \frac{1}{8}(4{k^2} + 12k + 9)\\ < \frac{1}{8}{(2k + 3)^2}\\ < \frac{1}{8}{[2(k + 1) + 1]^2}\end{array}$

Prove that $\displaystyle \frac{1}{{1 \cdot 3}} + \frac{1}{{3 \cdot 5}} + \frac{1}{{5 \cdot 7}} + \ldots \ldots + \frac{1}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}} = \frac{n}{{2n + 1}}$

By partial fraction, we know that

$\dfrac{1}{(2K-1)(2K+1)}=\dfrac{1}{2}(\dfrac{1}{2K-1}-\dfrac{1}{2K+1})$

Using Telescoping sum,

$\displaystyle\sum_{k=1}^{n}\dfrac{1}{(2K-1)(2K+1)}=\sum_{k=1}^{n}\dfrac{1}{2}(\dfrac{1}{2k-1}-\dfrac{1}{2k+1})$

$=\dfrac{1}{2}(1-\dfrac{1}{2n+1})$

$=\dfrac{n}{2n+1}$

Hence, proved.

Let $P(n)=n(n+1)(n+2)$ is divisible by $12$ then which one of the following is not true.
i) $P(3)$
ii) $P(4)$
ii) $P(5)$
iv) $P(8)$

Given,

$P(n)=n(n+1)(n+2)$ .

Now,

$P(3)=3.4.5=60$ is divisible by

$12$ .

ii)

$P(4)=4.5.6=120$ is divisible by

$12$ .

iii)

$P(5)=5.6.7=210$ not divisible by

$12$ .

iv)

$P(8)=8.9.10=720$ is divisible by

$12$ .

$P(5)$ is not true.

Using PMI, prove that ${3^{2n + 2}} - 8n - 9$ is divisible by $64$ .

Let $p\left( x \right)={{3}^{2n+2}}-8x-9$ is divisible by $64$ …..(1)

When put $n=1$ ,

$p\left( 1 \right)={{3}^{4}}-8-9=64$ which is divisible by $64$

Let $n=k$ and we get

$p\left( k \right)={{3}^{2k+2}}-8k-9$ is divisible by $64$

${{3}^{2k+2}}-8k-9=64m$ where $m\in N$ …..(2)

Now we shall prove that $p\left( k+1 \right)$ is also true

$p\left( k+1 \right)={{3}^{2\left( k+1 \right)+2}}-8\left( k+1 \right)-9$ is divisible by $64$ .

Now,

$p\left( k+1 \right)={{3}^{2\left( k+1 \right)+2}}-8\left( k+1 \right)-9={{3}^{2}}{{.3}^{2k+2}}-8k-17$

$={{9.3}^{2k+2}}-8k-17$

$=9\left( 64m+8k+9 \right)-8k-17$

$=9.64m+72k+81-8k-17$

$=9.64m+64k+64$

$=64\left( 9m+k+1 \right),$ Which is divisibility by $64$

Thus $p\left( k+1 \right)$ is true whenever $p\left( k \right)$ is true.

Hence, by principal mathematical induction,

$p\left( x \right)$ is true for all natural number $p\left( x \right)={{3}^{2n+2}}-8x-9$ is divisible by 64 $n\in N$

Use mathematical induction to prove:
$\dfrac{1}{{1 \cdot 3}} + \dfrac{1}{{3 \cdot 5}} + \dfrac{1}{{5 \cdot 7}} + \ldots + \dfrac{1}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}} = \dfrac{n}{{2n + 1}}$

$\dfrac{1}{1.3} + \dfrac{1}{3.5} + \dfrac{1}{5.7} + \cdots + \dfrac{1}{ (2n-1)(2n+1)} = \dfrac{n}{2n+1}$

$p(1) = \dfrac{1}{1.3} = \dfrac{1}{3}$ true

$p(n)$ is true for some $k \in N$

i.e $\dfrac{1}{1.3} + \dfrac{1}{3.5} + \cdots + \dfrac{1}{(2k-1)(2k+1)} = \dfrac{k}{2k+1}$

we need to prove $P(k+1)$ is true whenever $p(k)$ is true.

Now $\dfrac{1}{1.3} + \dfrac{1}{3.5} + \cdots + \dfrac{1}{(2k-1)(2k+1)} + \dfrac{1}{(2k+1)(2k+3)}$

$= [\dfrac{1}{1.3} + \dfrac{1}{3.5} + \cdots + \dfrac{1}{(2k-1)(2k+1)} ]+ \dfrac{1}{(2k+1)(2k+3)}$

$= \dfrac{k}{2k+1} + \dfrac{1}{(2k+1)(2k+3)}$

$= \dfrac{(2k+3)k + 1}{(2k+1)(2k+3)}$

$=\dfrac{ 2k^2+3k+1}{(2k+1)(2k+3)}$

$=\dfrac{k+1}{2k+3}$

Thus $p(k+1)$ is true.

@page { margin: 2cm } p { margin-bottom: 0.25cm; line-height: 120% }

Thus PMI $p(n)$ is true $\forall n \in N$

Prove using PMI: $\displaystyle \frac{1}{{1.2.3}} + \frac{1}{{2.3.4}} + \frac{1}{{3.4.5}} + ... + \frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{{n\left( {n + 3} \right)}}{{4\left( {n + 1} \right)\left( {n + 2} \right)}}$

$\begin{array}{l} P(n):\displaystyle \frac { 1 }{ { 1.2.3 } } +\frac { 1 }{ { 2.3.4 } } +\frac { 1 }{ { 3.4.5 } } +...+\frac { 1 }{ { n\left( { n+1 } \right) \left( { n+2 } \right) } } =\displaystyle \frac { { n\left( { n+3 } \right) } }{ { 4\left( { n+1 } \right) \left( { n+2 } \right) } } \\ for\, n=1,we\, have \\ P(1):\frac { 1 }{ { 1.2.3 } } =\displaystyle \frac { { 1\left( { 1+3 } \right) } }{ { 4\left( { 1+1 } \right) \left( { 1+2 } \right) } } =\displaystyle \frac { { 1.4 } }{ { 4.2.3 } } =\frac { 1 }{ { 1.2.3 } } ,which\, is\, true \\ Let,P(k)be\, true\, for\, some\, positive\, integer\, k,i.e., \\ P(k):\displaystyle \frac { 1 }{ { 1.2.3 } } +\frac { 1 }{ { 2.3.4 } } +\frac { 1 }{ { 3.4.5 } } +...+\frac { 1 }{ { k\left( { k+1 } \right) \left( { k+2 } \right) } } =\frac { { k\left( { k+3 } \right) } }{ { 4\left( { k+1 } \right) \left( { k+2 } \right) } } .........(1) \\ we\, shall\, now\, prove\, that\, P(k+1)is\, true \\ \\ consider \\ P(k+1):\displaystyle \left[ { \frac { 1 }{ { 1.2.3 } } +\frac { 1 }{ { 2.3.4 } } +\frac { 1 }{ { 3.4.5 } } +...+\frac { 1 }{ { k\left( { k+1 } \right) \left( { k+2 } \right) } } } \right] +\frac { 1 }{ { \left( { k+1 } \right) \left( { k+2 } \right) \left( { k+3 } \right) } } \\ =\displaystyle \frac { { k\left( { k+3 } \right) } }{ { 4\left( { k+1 } \right) \left( { k+2 } \right) } } +\frac { 1 }{ { \left( { k+1 } \right) \left( { k+2 } \right) \left( { k+3 } \right) } } .................u\sin g(1) \\ =\displaystyle \frac { 1 }{ { \left( { k+1 } \right) \left( { k+2 } \right) } } \left\{ { \frac { { k\left( { k+3 } \right) } }{ 4 } +\frac { 1 }{ { \left( { k+3 } \right) } } } \right\} \end{array}\\$

$\begin{array}{l} =\displaystyle \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left\{ {\frac{{k{{\left( {k + 3} \right)}^2} + 4}}{{4\left( {k + 3} \right)}}} \right\}\\ =\displaystyle \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left\{ {\frac{{k\left( {{k^2} + 9 + 6k} \right) + 4}}{{4\left( {k + 3} \right)}}} \right\}\\ =\displaystyle \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left\{ {\frac{{{k^3} + 9k + 6{k^2} + 4}}{{4\left( {k + 3} \right)}}} \right\}\\ =\displaystyle \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left\{ {\frac{{{k^3} + 2{k^2} + k + 4{k^2} + 8k + 4}}{{4\left( {k + 3} \right)}}} \right\}\\ = \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left\{ {\dfrac{{k({k^2} + 2k + 1) + 4({k^2} + 2k + 1)}}{{4\left( {k + 3} \right)}}} \right\}\\ =\displaystyle \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left\{ {\dfrac{{k{{(k + 1)}^2} + 4{{(k + 1)}^2}}}{{4\left( {k + 3} \right)}}} \right\}\\ = \dfrac{{{{(k + 1)}^2}(k + 4)}}{{4\left( {k + 3} \right)\left( {k + 1} \right)\left( {k + 2} \right)}}\\ = \dfrac{{(k + 1)\left\{ {(k + 1) + 3} \right\}}}{{4\left\{ {(k + 1) + 1} \right\}\left\{ {(k + 1) + 2} \right\}}}\\Thus,P(k + 1),is\,true\,whenever\,P(k)\,is\,true\end{array}$

Prove : $1 + 2 + 3 + ... + n = \dfrac{1}{2} n (n + 1)$

The series

$1+2+3+.......+n$ is in A.P.

where

first term,

$a=1$

common difference,

$d=1$

last term,

$l=n$

No. of terms

$=n$

In A.P.

Sum of first

$n$ terms,

$S_{n}=\frac{n}{2}\left ( a+l \right )$

$=\frac{n}{2}\left ( 1+n \right )$

$1.2 + 2.3 + 3.4 + .. + n (n + 1) = \dfrac{1}{m} n (n + 1) (n + 2)$

$n$ th term of the given series is,

$a_{n}=n(n+1)$

$a_{n}=n^{2}+n$

Sum of

$n$ terms is

$S_{n}=\sum_{k=1}^{n}k^{2}+\sum_{k=1}^{n}k$

$=\cfrac{n(n+1)(2n+1)}{6}+\cfrac{n(n+1)}{2}$

$=\cfrac{n(n+1)}{2}\left [ \cfrac{2n+1}{3}+1 \right ]$

$=\cfrac{n(n+1)}{2}\cdot \cfrac{2n+4}{3}$

$=\cfrac{n(n+1)(n+2)}{3}$

$\Rightarrow \cfrac{1}{m}=\cfrac{1}{3}$

$m=3$

Prove that by mathematical induction ${2^{3n}} - 1$ is divisible by $7$ for all natural numbers.

$P(n):2^{3n}-1$

If It's divisible by

$7$ then,

Base case:

$n=1$ then,

$P(n):2^{3n}-1=2^{3\times1}-1=7$

$n=k$ is true for some natural numbers, then It should also be true for

$(k+1)$ ,

For

$n=k$ ,

$P(k):2^{3k}-1=7d$

$P(k):2^{3k}=7d+1$ .....(1)

Where

$d=1,2,3,4.......so\ on$

Now we check for

$n=k+1$ ,

$P(k+1):2^{3(k+1)}-1=7d$

$P(k+1):2^{3k+3}=7d+1$

(Since

$a^{xy}=a^x a^y$ )

$P(k+1):2^{3k}2^3=7d+1$

Substituting value from equation (1),

$P(k+1):(7d+1)\times8=7d+1$

$56d+8=7d+1$

$7\times(8d+1)=7d$

So, it is true for both

$k\ and\ (k+1)$ .

Hence, from the principle of mathematical induction,

$2^{3n}-1$ is divisible by

$7$ for all natural numbers.

$2 + 4 + 6 + ... + 2n = m^2\times n (n + 1)$

$2+4+6+................+2n = m^2 \times n(n+1)$

$2(1+2+3+.............. +n) = m^2 \times n(n+1)$

$2\times \cfrac{n(n+1)}{2} = m^2 \times n(n+1)$ ( Sum of first

$n$ natural numbers

$= \cfrac{n(n+1)}{2}$ )

$n(n+1) = m^2 \times n(n+1)$

$\therefore m^{2}=1$

$\therefore m=\pm1$

Prove that product of two consecutive natural number cannot be $441$ .

$let\>the\>natural\>number\>be\>n\>then\>n(n+1)=441\\n^2+n-441=0\\D=b^2-4ac\\=\>1^2-4\times\>1\>\times\>(-441)\\=1+4\times\>441\\=1+(42)^2\\\therefore\>\sqrt\>D\>will\>not\>be\>an\>integer\\\>and\>hence\>n\>=(\frac{-b\>\pm\>\sqrt\>D}{2a})\>will\>not\>be\>\>an\>integer,\\\>so\>we\>can\>conclude\>product\>\\of\>two\>consecutive\>number\>can\>not\\\>be\>441$

$1.3 + 2.4 + 3.5 + ... + n (n + 2) = \dfrac{1}{m} n (n + 1)(2n + 7)$

$n$ th term of the given series is,

$a_{n}=n(n+2)$

$a_{n}=n^{2}+2n$

Sum of

$n$ terms is

$S_{n}=\sum_{k=1}^{n}k^{2}+2\sum_{k=1}^{n}k$

$=\cfrac{n(n+1)(2n+1)}{6}+2\times \cfrac{n(n+1)}{2}$

$=\cfrac{n(n+1)(2n+1)}{6}+n(n+1)$

$=n(n+1)\left [ \cfrac{2n+1}{6}+1 \right ]$

$=n(n+1)\left [ \cfrac{2n+7}{6} \right ]$

$=\cfrac{n(n+1)(2n+7)}{6}$

$\Rightarrow \cfrac{1}{m}=\cfrac{1}{6}$

$m=6$

Prove that : $1 + 3 + 3^2 + ... + 3^{n-1} = \dfrac{(3^n - 1)}{2}$

The given series

$1+3+3^{2}+......+3^{n-1}$ is in G.P. where

first term,

$a_{1}=1$

common ratio,

$r=\cfrac{a_{2}}{a_{1}}=\cfrac{3}{1}=3$

$n$ th term,

$a_{n}=a_{1}r^{n-1}=3^{n-1}$

Sum of

$n$ terms,

$S_{n}=\cfrac{a\left ( r^{n}-1 \right )}{r-1}$

$=\cfrac{1\left ( 3^{n}-1\right )}{3-1}$

$=\cfrac{3^{n}-1}{2}$

Prove $1.3 + 3.5 + 5.7 + ...+ (2n -1)(2n + 1) = \dfrac{n(4n^2 + 6n -1)}{3}$

$k$ th term of the series is

$=(2k-1)(2k+1)$

$=4k^{2}-1$

Sum of first

$n$ terms is,

$S_{n}=4\sum_{k=1}^{n}k^{2}-\sum_{k=1}^{n}1$

$=\cfrac{4n(n+1)(2n+1)}{6}-n$

$=\cfrac{4n^{3}+6n^{2}-n}{3}$

$=\cfrac{n\left ( 4n^{2}+6n-1 \right )}{3}$

Prove that $1^2 + 2^2 + 3^2 + .... + n^2 > \dfrac{n^2}{3}$ for all $n \in N$ using principle of mathematical induction.

Formula is given by,

$1^2+2^2+3^2+......+n^2=\dfrac{n(n+1)(2n+1)}{6}$

To prove:

$1^2+2^2+3^2+......+n^2> \dfrac{n^2}{3}$

But,

$1^2+2^2+3^2+......+n^2=\dfrac{n(n+1)(2n+1)}{6}$

$\Rightarrow$ To prove

$\dfrac{n(n+1)(2n+1)}{6}>\dfrac{n^2}{3}$

If we put

$n=1$ , we get,

$\dfrac{1(1+1)(2+1)}{6}>\dfrac{1^2}{3}$

$1>\dfrac{1}{3}$

$\Rightarrow 1^2+2^2+3^2+......+n^2> \dfrac{n^2}{3}$

Prove by PMI:
$cos \theta + cos 2\theta + ..... + cosn \theta = sin\dfrac{n\theta}{2}.cos \theta c\dfrac{\theta}{2}.cos\dfrac{(n+1)\theta}{2}$

$\begin{array}{l} \cos \theta +\cos 2\theta +\cos 3\theta +...........\cos n\theta \\ { e^{ i\theta } }=\cos \theta +i\sin \theta \\ \sum _{ k=1 }^{ n }{ \cos k\theta =R }\sum _{ k=1 }^{ n }{ { e^{ ki\theta } } } \\ =R.{ e^{ i\theta } }\left( { 1+{ e^{ i\theta } }+{ e^{ 2i\theta } }+.....{ e^{ \left( { n-1 } \right) i\theta } } } \right) \\ =R\left[ { { e^{ i\theta } }.\left( { \frac { { { e^{ i\theta } }-1 } }{ { { e^{ i\theta } }-1 } } } \right) } \right] \\ =R\left[ { { e^{ i\theta } }.\frac { { { e^{ \frac { \theta }{ 2 } } }\left( { 2i\sin \frac { { n\theta } }{ 2 } } \right) } }{ { { e^{ \frac { { i\theta } }{ 2 } } }.2i\sin \frac { \theta }{ 2 } } } } \right] \\ =R\left[ { { e^{ \frac { { i\theta \left( { n+1 } \right) } }{ 2 } } }.\frac { { \sin n\frac { \theta }{ 2 } } }{ { \sin \frac { \theta }{ 2 } } } } \right] \\ =R\left[ { \cos \left( { n+1 } \right) \frac { \theta }{ 2 } +i\sin \left( { n+1 } \right) \frac { \theta }{ 2 } .\frac { { \sin n\frac { \theta }{ 2 } } }{ { \sin \frac { \theta }{ 2 } } } } \right] \\ =\frac { { \sin n\frac { \theta }{ 2 } } }{ { \sin \frac { \theta }{ 2 } } } .\cos \left( { n+1 } \right) \frac { \theta }{ 2 } \\ =\sin \frac { { n\theta } }{ 2 } .\cos ec\frac { \theta }{ 2 } .\cos \left( { n+1 } \right) \frac { \theta }{ 2 } \\Hence\ Proved. \end{array}$

For any odd integer $n\geq 1$ , Find the value of
$n^3-(n-1)^3+...........+(-1)^{n-1}1^3$ .

1329156_1080968_ans_ec95dad172344e7d9612914b90d75c2b.jpg

Prove by Induction $a+ar+{ ar }^{ 2 }+.....$ up to n terms = $\cfrac { a\left( { r }^{ n }-1 \right) }{ r-1 } ,r\neq 1$

consider the base case

$n=1$

$LHS = a,\,RHS = a\left(\dfrac{r^{1}-1}{r-1}\right)=a$

Hence, statement holds for

$n=1$

Now assume statement is valid for

$n=k$

i.e

$a+ar+ar^{2}.......ar^{k-1}=a\left(\dfrac{r^{k}-1}{r-1}\right)$

Adding

$ar^{k}$ to both sides, we get

$a+ar+ar^{2}........ar^{k}=a\left(\dfrac{r^{k}-1}{r-1}+\dfrac{r^{k+1}-r^{k}}{r-1}\right)$

$a+ar+ar^{2}........ar^{k}=a\left(\dfrac{r^{k+1}-1}{r-1}\right)$

i.e statement holds for

$n=k+1$ also

Hence, by Mathematical function,

Statement is true for all

$n \epsilon N$

Prove that ${25^n} - {20^n} - {8^n} + {3^n}$ is divisible by $5$ .

Consider the given expression, ${{8}^{n}}-{{3}^{n}}$ .

We to prove ${{8}^{n}}-{{3}^{n}}$ is divisible by $5.$

We know that,

$[{{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left( {{a}^{n-1}}+{{a}^{n-2}}.b+{{a}^{n-3}}{{b}^{2}}+{{a}^{n-4}}{{b}^{3}}+\,............+{{b}^{n}}^{-1} \right)]$

Put $a=8,b=3,$ we get

${{8}^{n}}-{{3}^{n}}=\left( 8-3 \right)\left( {{8}^{n-1}}+{{8}^{n-2}}.3+{{8}^{n-3}}{{3}^{2}}+{{8}^{n-4}}{{3}^{3}}+\,............+{{3}^{n}}^{-1} \right)$

${{8}^{n}}-{{3}^{n}}=5\left( {{8}^{n-1}}+{{8}^{n-2}}.3+{{8}^{n-3}}{{3}^{2}}+{{8}^{n-4}}{{3}^{3}}+\,............+{{3}^{n}}^{-1} \right)$

Which is divisible by $5.$ hence proved.

Prove that if either of $2a+3b$ and $9a+5b$ is divisible by 17 ,so is the other , $a,b \in N$

Let

$2a+3b= 17k$

Then

$a= \dfrac{(17k-3b)}{2}$ ----- (i)

Consider,

$9a+5b$

$=\dfrac{(153k-27b)}{2}+5b$ ---from (i)

$=\dfrac{(153k-17b)}{2}$

$= \dfrac{17(9k-b)}{2}$

which is divisible by 17

and vice versa

Solve $\dfrac{1}{{1.2.3}} + \dfrac{1}{{2.3.4}} + \dfrac{1}{{3.4.5}} + \cdots + \dfrac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} = \dfrac{{n\left( {n + 3} \right)}}{{4\left( {n + 1} \right)\left( {n + 2} \right)}}$

Let us assume that,

$p\left( x \right) = \dfrac{1}{{1.2.3}} + \dfrac{1}{{2.3.4}} + \dfrac{1}{{3.4.5}} + \cdots + \dfrac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} = \dfrac{{n\left( {n + 3} \right)}}{{4\left( {n + 1} \right)\left( {n + 2} \right)}}$

for

$n=1$

L.H.S

$=\dfrac{1}{{1.2.3}} = \dfrac{1}{6}$

R.H.S

$= \dfrac{{1.\left( {1 + 3} \right)}}{{4\left( {1 + 1} \right)\left( {1 + 2} \right)}} = \dfrac{1}{6}$

$P(1)$ is true.

Let,

$P(k)$ is true

i.e,

$= \dfrac{1}{{1.2.3}} + \dfrac{1}{{2.3.4}} + \dfrac{1}{{3.4.5}} + \cdots + \dfrac{1}{{k\left( {k + 1} \right)\left( {k + 2} \right)}} = \dfrac{{k\left( {n + 3} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)}}$

To show :-

$P(k+1)$ is true

Now,

$= \dfrac{1}{{1.2.3}} + \dfrac{1}{{2.3.4}} + \cdots + \dfrac{1}{{k\left( {k + 1} \right)\left( {k + 2} \right)}} + \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}$

$= \dfrac{{k\left( {k + 3} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)}} + \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}$

$= \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left[ {\dfrac{{k\left( {k + 3} \right)}}{4} + \dfrac{1}{{k + 3}}} \right]$

$= \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\,\dfrac{{k{{\left( {k + 3} \right)}^2} + 4}}{{4\left( {k + 3} \right)}}$

$= \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\,\,\dfrac{{k\left( {{k^2} + 6k + 9} \right) + 4}}{{4\left( {k + 3} \right)}}$

$= \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\,\,\dfrac{{{k^3} + 6{k^2} + 9k + 4}}{{4\left( {k + 3} \right)}}$

$= \dfrac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\,\,\dfrac{{{k^3} + {k^2} + 5{k^2} + 4k4}}{{4\left( {k + 3} \right)}}$

$= \,\dfrac{{{k^2}\left( {k + 1} \right) + 5k\left( {k + 1} \right) + 4\left( {k + 1} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}$

$=\dfrac{{\left( {k + 1} \right)\left( {{k^2} + 5k + 4} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}$

$= \dfrac{{{k^2} + 4k + k + 4}}{{4\left( {k + 2} \right)\left( {k + 3} \right)}}$

$= \dfrac{{k\left( {k + 4} \right) + \left( {k + 4} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)}}$

$= \dfrac{{\left( {k + 1} \right)\left( {k + 4} \right)}}{{4\left( {k + 2} \right)\left( {k + 3} \right)}}$ Hence proved.

Suppose $m,n$ are integers and $m=n^2-n$ . Then show that $m^2-2m$ is divisible by $24$ .

Given

$:m={n}^{2}-n$

$\Rightarrow\,{m}^{2}-2m={\left({n}^{2}-n\right)}^{2}-2\left({n}^{2}-n\right)$

$={n}^{4}+{n}^{2}-2{n}^{3}-2{n}^{2}+2n$

$={n}^{4}-2{n}^{3}-{n}^{2}+2n$

$={n}^{3}\left(n-2\right)-n\left(n-2\right)$

$=\left({n}^{3}-n\right)\left(n-2\right)$

$=n\left({n}^{2}-1\right)\left(n-2\right)$

$=n\left(n-1\right)\left(n+1\right)\left(n-2\right)$

$=\left(n-2\right)\left(n-1\right)n\left(n+1\right)$

The above expression represents the product of

$4$ consecutive numbers.

$\therefore\,$ it should be divisible by

$24$ provided

$n>2$ .

If $A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \end{bmatrix}$ , prove that

$A^n=\begin{bmatrix} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix}, n\in N$ .

Given

$A=\begin{bmatrix} 1& 1& 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$

To prove :

$A^n=\begin{bmatrix} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix}, n\in N$ .

Proof :

Let P(n) be the given statement.

For

$n=1$ ,

$P(1)=A=\begin{bmatrix} 3^0 & 3^0 & 3^0 \\ 3^0 & 3^0 & 3^0 \\ 3^0 & 3^0 & 3^0 \end{bmatrix} =\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}=A$

Therefore, statement is true for

$n=1$ .

Now, let us assume that statement is true for

$n=k$ .

$P(k):A^k=\begin{bmatrix} 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \end{bmatrix}$

Now, we shall prove the statement for

$n=k+1$ , we have to show,

$P(k+1)=A^{k+1}=\begin{bmatrix} 3^k & 3^k & 3^k \\ 3^k & 3^k & 3^k \\ 3^k & 3^k & 3^k \end{bmatrix}$

LHS =

$A^{k+1}=A^k.A$

$=\begin{bmatrix} 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \end{bmatrix} \begin{bmatrix} 1&1&1 \\ 1&1&1\\1&1&1\end{bmatrix}$

$=\begin{bmatrix} 3^k & 3^k & 3^k \\ 3^k & 3^k & 3^k \\ 3^k & 3^k & 3^k \end{bmatrix}$ = RHS

Statement is true for

$n=k+1$ .

Hence, by principal of mathematical induction, statemebt is true for all n, where

$n\in N$ .

Prove that $1.2 + 2.3 +.......n(n+1)$ = $\frac{{n(n + 1)(n + 2)}}{3}$ in mathematical induction.

We shall prove the result by principle of mathematical induction.

checking for

$n = 1$ ,

$LHS : 1.2 = 2$

$RHS : \dfrac{1}{3} \times 1 \times 2 \times 3 = 2$ .

Hence true for

$n = 1$

Let us assume the result is true for

$n = k$ ie.,

$1.2 + 2.3 +.....k(k+1) = \dfrac{1}{3} \times k \times (k+1) \times (k+2)$

We shall prove the result to be true for

$n = k+1$ .

that is, to prove

$1.2 + 2.3 .....+ k(k+1) + (k+1)(k+2) = \dfrac{1}{3} (k+1) (k+2) (k+3)$

consider

$LHS : 1.2 + 2.3 .....+ k(k+1) + (k+1)(k+2)$

$= \dfrac{1}{3} \times k \times (k+1) \times (k+2) + (k+1)(k+2)$

$=(k+1)(k+2) [\dfrac{1}{3}(k + 1)]$

$=(k+1)(k+2)(k+3)\dfrac{1}{3}$

$=RHS$ .

Hence the result holds for

$n=k+1$ .

Hence proof is complete by PMI and therefore the result holds.

Prove that $2.7^{n}+3.5^{n}-5$ is divisible by $24$ , for all $n\in N$ .

$2(7)^{m}+3(5)^{m}-5=P(n)$

divisible by

$24=4\times$ natural no

for

$n=1$

$P(n)=$ true for

$n=1$

$P(k+1)=27^{k+1}+35^{k+1}-5$

$\Rightarrow 2.2\times 7^{k}+5.3.5^{k}-5=24\ m$

$\Rightarrow 7\times 24\ m-6.5^{k}+30$

$\Rightarrow 7\times 24\ m-6(5^{k}-5)$

$(5^{k}-5)$ is a multiple of

$4$

$=7\times 24\ m-6(4p)$

$P$ is a natural No

$=7\times 24\ m-24P$

$=24(7\ m-P)=24\times r$

$r=7\ m=P$ is some natural No

$P(k+1)$ is true whenever

$P(k)$ is true.

Prove by method of induction, for all $n \in N$
$2 + 4 + 6 + .... + 2n = n\left( {n + 1} \right)$

First, that answer is correct

$2+4+b+... +2n = 2 (1+2+3....+n)= \dfrac{n (n+1)}{2}$

$= n (n+1)$

Second, suppose that is true you have

$2+4+b+.....+2n$

$=2 \times (n+1)$

That is

$S(n)$ and

$2(n+1)$ to both sides giving

$2+4+.....+2n+2 (n+1)= n* (n+1)+2 (n+1)(n+1)$

Which is

$S(n+1)$ Hence proved.

that

$2+4+b+......+ 2n= n(n+1)$

prove the inequalities ${\left( {n!} \right)^2} \le {n^n}\left( {n!} \right) < \left( {2n} \right)!$ for all positive integers n .

$\\consider$

$(n!)^2\leq n^n(n!) part$

$\\n\leq n$

$\\(n-1)\leq n$

$\\(n-2)\leq n$

$\\.....$

$\\1\leq n$

$\\\therefore n(n-1)(n-2)......1\leq n.n.n......n\>times$

$\\n!\leq n^n$

multiply both sides by n!, we get:

$\\(n!)^2\leq n^n(n!)$

Now consider the second part $n^n(n!)\leq (2n!)$

$\\as$

$\\ n\leq 2n$

$\\n\leq (2n-1)$

$\\n\leq (2n-2)$

$\\.....$

$\leq [2n-(n-1)]$

$\therefore n.n..... n\>times \leq 2n(2n-1)(2n-2)........[2n-(n-1)]$

$n^n\leq 2n(2n-1)(2n-2)........[n+1]$

multiply both sides by n!, we get

$\\n^n n! \leq 2n(2n-1)(2n-2)........[n+1](n!)$

$\\n^n n! \leq(2n!)$

Hence Proved!

Let $P(n)$ be the statement $3^{n} > n^{n}$ . If $P(n)$ is true, prove that $P(n+1)$ is true.

$P(n):3^n>n^n$

for

$n=1$

$P(1):3'1>1^1$

$3>1$

$\therefore$ It is true for

$n=1$

$P(K)\Rightarrow 3^k>k^k$

It is given that

$P(h)$ is true, so

$P(k)$ is true for

$n=k$

$P(k+1)=3^{k+1}> (k+1)^{k+1}$

$3^k.3^1> (k+1)^k(k+1)$

By the principle for mathematics production,

$P(n+1)$ is true, when

$P(n)$ is true

$\therefore P(n+1)$ is true.

Show that $1^{2}+(1^{2}+2^{2})+(1^{2}+2^{2}+3^{2})+$ .... up to $n$ terms $=\dfrac{n(n+1)^{2}}{12}$ , $n\epsilon N$

1055337_1114325_ans_1661be2846ee4b7283e01d3ddad4dc19.png

${7^{2n}} + {2^{3n - 3}}{.3^{n - 1}}$ is divisible by $25$ for any natural number $n \ge 1$ . Prove that by mathematical

Step 1. Check if

$P(i)$ is correct

We get

$P(1)=7^2+2^0\times 3^0=49+1=50$

Which is divisible by

$25$

$P(1)$ is true.

Step 2. Let

$P(K)$ is true, which means

$7^{2k}+3^{3k-3}3^{k-1}=25m$

When

$m=integer$

Now check if

$P(K+1)$ is True

Hence,

$P(KH)=7^{2(K+H)}+2^{3(K+1)-3}\times 3^{(K+1)-1}$

$=(7^{2k}\times 49)+2^{(3k-3)}\times 8\times 3^{(k-1)}\times 3$

$=49\times 7^{2k}+24[2^{3k-3}\,\,3^{k-1)} ]$

$=49\times 7^{2k}+24[25m-7^{2k}]$

$=49.7^{2k}-24.7^{2k}+24.25m$

$=25[7^{2k}+24m]$

$=25\times$ some integers value

$\therefore$

$P(K+1)$ is true whenerver

$P(K)$ is true.so using

$PMI$ , we can conclude

$P(n)$ is true for all

$n \in N$

What must be added to ${x^4} + 2{x^3} - 2{x^2} + x - 1$ . So that result is exactly divisible by ${x^2} + x - 2$ ?

${ x }^{ 4 }+2{ x }^{ 3 }-2{ x }^{ 2 }+{ x }-{ 1 }\\ ={ x }^{ 4 }+{ x }^{ 3 }+{ x }^{ 3 }-2{ x }^{ 2 }+{ x }^{ 2 }-{ x }^{ 2 }+{ x }{ +1 }\\ =({ x }^{ 4 }+{ x }^{ 3 }-2{ x }^{ 2 })+({ x }^{ 3 }+{ x }^{ 2 }-{ 2x })-({ x }^{ 2 }+{ x }-{ 2 })+({ 4x }-{ 3 })\\ ={ x }^{ 2 }({ x }^{ 2 }+{ x }-{ 2 })+{ x }({ x }^{ 2 }+{ x }-{ 2 })-({ x }^{ 2 }+{ x }-{ 2 })+({ 4x }-{ 3 })\\$

Here we have to remove the last bracket terms

${4x}-{3}$ so that the above expression is divisible by the expression given in question

So we have to add

${3}-{4x}$ to make the expression divisible

From principle of mathematical induction prove that
$1+2+3+....+n=\dfrac{1}{2}n(n+1)$

$1+2+3\quad n$

$AP$

Sum of

$n$ term

$=\dfrac {n}{2}[2a+(n-1)d]$

$=\left (\dfrac {n}{2}\right) [2(a) +(n-1)1]$

$=\dfrac {n}{2}[n+1]$

If $33!$ is divisible by $2^n$ , then find the maximum value of n.

For prime number ${p}$ and ${n}$ the highest power of ${p}$ that can divide ${p!}$ can be given as

$\sum { \left\lfloor \dfrac { n }{ { p }^{ k } } \right\rfloor } \\ { n }={ 33 }\quad ,\quad { p }={ 2 }\\ =\left\lfloor \dfrac { 33 }{ { 2 }^{ 1 } } \right\rfloor +\left\lfloor \dfrac { 33 }{ { 2 }^{ 2 } } \right\rfloor +\left\lfloor \dfrac { 33 }{ { 2 }^{ 3 } } \right\rfloor +\left\lfloor \dfrac { 33 }{ { 2 }^{ 3 } } \right\rfloor +\left\lfloor \dfrac { 33 }{ { 2 }^{ 4 } } \right\rfloor +\left\lfloor \dfrac { 33 }{ { 2 }^{ 5 } } \right\rfloor \\ ={ 16 }+{ 8 }+{ 4 }+{ 2 }+{ 1 }\\ ={ 31 }$

So ${n}={31}$

If $A=\begin{bmatrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{bmatrix}$ then prove that ${ A }^{ n }=\begin{bmatrix} \cos { n\theta } & \sin { n\theta } \\ -\sin { n\theta } & \cos { n\theta } \end{bmatrix}$ , $n\in N$ .

$A=\begin{bmatrix} \cos \theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix}$

$\therefore A^2=\begin{bmatrix} \cos \theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$

$\therefore A^2=\begin{bmatrix} \cos^2-\sin^2\theta & \cos\theta \sin\theta +\sin\theta \cos\theta\\ -\sin\theta \cos\theta -\sin\theta \cos\theta & -\sin^2\theta +\cos^2\theta\end{bmatrix}$

$\therefore A^2=\begin{bmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta\end{bmatrix}$

$\therefore A^3=\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{bmatrix}\begin{bmatrix} \cos\theta & \sin \theta\\ -\sin\theta & \cos\theta\end{bmatrix}$

$\therefore A^3=\begin{bmatrix} \cos 2\theta \cos\theta -\sin 2\theta \sin\theta & \sin\theta \cos 2\theta +\sin 2\theta \cos \theta\\ -\sin 2\theta \cos\theta -\sin\theta \cos 2\theta & -\sin \theta \sin\theta +\cos 2\theta\cos \theta\end{bmatrix}$

$\therefore A^3=\begin{bmatrix} \cos 3\theta & \sin 3\theta\\ -\sin 3\theta & \cos 3\theta\end{bmatrix}$

$\therefore A^n=\begin{bmatrix} \cos n\theta & \sin n\theta\\ -\sin n\theta & \cos n\theta\end{bmatrix}$

$n\in R$ .

1072179_1160175_ans_e1f463162a004b418fc9e8782c12e1e1.png

$(n+1)(n+2)$ is an even number

$\Rightarrow (n+1)(n+2)\rightarrow$ is an even NO

$\Rightarrow n^2+2n+2$

$n\Rightarrow$ even

$\Rightarrow$ odd

$\times$ even=even

$n=$ odd

$\Rightarrow$ even

$\times$ odd=even

Show that any positive odd integers is of the form $6q+1,6q+3$ or $6q+5$ , where $q$ is some integer.

Let take a as any positive integer and b=6

Then using Euclid's algorithm we get

$a=6q+r$

here

$r$ is remainder and value of

$q\ge0$

and

$r=0,1,2,3,4,5$ because

$0\le r<b$ and the value

$b=6$

So total possible forms will be

$6q+0,\ 6q+1,\ 6q+2,\ 6q+3,\ 6q+4,\ 6q+5$

$6q+0$

6 is divisible by 2 so it is even number

$6q+1$

6 is divisible by 2 but 1 is not divisible by 2 so it is odd number

$6q+2$

6 is divisible by 2 and 2 is also divisible by 2 so it is even number

$6q+3$

6 is divisible by 2 but 3 is not divisible by 2 so it is odd number

$6q+4$

6 is divisible by 2 and 4 is also divisible by 2 so it is even number

$6q+5$

6 is divisible by 2 but 5 is not divisible by 2 so it is odd number

So, odd numbers will in form of

$6q+1,\ 6q+3,\ 6q+5.$

Use the principle of mathermatical induction to prove that
${1}^{3}+{2}^{3}+.....{n}^{3}=\cfrac{1}{4}{n}^{2}{(n+1)}^{2}$ for every natural number $n$ .

Let

$P(n) : 1^3+2^3+3^3+......+n^3=\left(\dfrac{n(n+1)}{2}\right)^2$

For

$n=1$

$LHS=1^3=1$

$RHS=\left(\dfrac{1(1+1)}{2}\right)^2=1$

Hence

$LHS=RHS$

$\therefore P(n)$ is true for

$n=1$

Assume that P(k) is true

$1^3+2^3+3^3+....+k^3=\left(\dfrac{k(k+1)}{2}\right)^2......(1)$

We will prove that P(k+1) is true

$1^3+2^3+3^3+....+k^3=\left(\dfrac{(k+1)(k+1+1)}{2}\right)^2$

$1^3+2^3+3^3+....+k^3=\left(\dfrac{(k+1)(k+2)}{2}\right)^2......(2)$

We have to prove P(k+1) from P(k) i.e (2) from (1)

From (1)

$1^3+2^3+3^3+.....+k^3=\left(\dfrac{k(k+1)}{2}\right)^2$

Adding

$(k+1)^3$ both sides,

$1^3+2^3+3^3+.....+k^3+(k+1)^3=\left(\dfrac{k(k+1)}{2}\right)^2+(k+1)^3$

$=\dfrac{k^2(k+1)^2}{4}+(k+1)^3$

$=\dfrac{k^2(k+1)^2+4(k+1)^3}{4}$

$=\dfrac{(k+1)^2(k^2+4k+4)}{4}$

$=\dfrac{(k+1)^2(k^2+2k+2k+4)}{4}$

$=\dfrac{(k+1)^2(k+2)(k+2)}{4}$

$=\dfrac{(k+1)^2(k+2)^2}{4}$

$=\left(\dfrac{(k+1)(k+2)}{2}\right)^2$

Thus,

$1^3+2^3+3^3+....+k^3=\left(\dfrac{(k+1)(k+2)}{2}\right)^2$ i.e P(k+1) is true whenever P(k) is true.

$\therefore$ By the principle of mathematical induction, P(n) is true for n, where n is a natural number.

Given an example of a statement $P(n)$ which is true for all $n\ge 4$ but $P(1), P(2)$ and $P(3)$ are not true. Justify your answer.

Consider the statement

$P(n):3n < n!$

For

$n=1,\ 3\times1<1!,$ Which is not true

For

$n=2,\ 3\times2<2!,$ Which is not true

For

$n=3,\ 3\times3<3!,$ Which is not true

For

$n=4,\ 3\times4<4!,$ Which is true

For

$n=5,\ 3\times5<5!,$ Which is true

Prove that $n^{2}-n$ in divisible by $2$ for any positive length $n$ .

Let,

$f(n)=n^2-n$ ;

$n\in N$

Now,

$f(1)=0$ is divisible by

$'2'$

Again

$f(2)=2^2-2=2$ ; divisible by

$'2'$

So,

$f(1), f(2)$ are true.

Let us assume that

$f(k)$ is true i.e.,

$f(k)=k^2-k=2k_1$ ;

$k_1\in z$

Now,

$f(k+1)=(k+1)^2-(k+1)$

$=k(k+1)$

$=k^2+k$

$=k^2-k+2k$

$=2k_1+2k$ [using

$(1)$ ]

$=2(k_1+k)$ ;

So,

$f(x+1)$ is divisible by

$2$ .

Now, by principle of mathematical induction we have,

$f(x)$ is true

$\forall n\in N$ .

1175748_1195490_ans_e81adfd3389146c6aaf880a32026d3ff.jpg

Prove that $1+2+3+.....+n=\dfrac{n(n+1)}{2}$ . for $n$ being a natural numbers.

Let,

$p(n)=1+2+3+...…+n$

Now,

$p(1)=1=\dfrac{1.(1+1)}{2}=1$

So,

$p(1)$ is true

Let, us assume that

$p(k)$ is true that is

$1+2+...…+k=\dfrac{k(k+1)}{2}$ ………

$(1)$

We shall prove that

$p(k+1)$ is true

Now,

$p(k+1)$

$=(1+2+.....+k)+(k+1)$

$=\dfrac{k(k+1)}{2}+(k+1)$

$=\dfrac{(k+1)(k+2)}{2}$

$=\dfrac{(k+1)(k+1+1)}{2}$

So,

$p(k+1)$ is true.

Now by principle of mathematical induction curve

$p(n)$ is true

$\forall n\in N$ .

1175715_1194770_ans_1f1a5dcce95f4dada6e6aef963dfb223.jpg

If $(27)^{999}$ is divided by 7, then the remainder is :

$(27)^{999}=(28-1)^{999}$

$={^{999}C_0}(28)^{999}-{^{999}C_1}(28)^{998}….+{^{999}C_{998}}(28)-{^{999}C_{999}}$

$=28\left[{^{999}C_0}(28)^{998}-{^{999}C_1}(28)^{997}……+{^{999}C_{998}}\right]-{^{999}C_{999}}$

$(27)^{999}=\dfrac{28\left[{^{999}C_0}(28)^{998}-{^{999}C_1}(28)^{997}…….+{^{999}C_{998}}\right]}{7}-\dfrac{1}{7}$

So remainder is remainder of

$\dfrac{-1}{7}$

which is

$6$ (because

$-1=7(-1)+6$ )

(remainder is always positive so we will not consider

$-1$ ).

1188157_1194969_ans_69bb322de05e45c0a1efe033f04cca67.jpg

Prove by the principle of mathematical induction that for all $n \in N$ :
$1 + 4+ 7 +.... + (3n - 2) = \dfrac{1}{2}n (3n - 1)$

The statement S1 : 1 =

$\dfrac{1}{2}(3−1) 2= 1$ is true.

Assume that Sk : 1 + 4 + 7 + . . . + (3k − 2) =

$\dfrac{k}{2}(3k−1)$ is true --(1)

and to prove that Sk+1 : 1 + 4 + 7 + . . . + (3(k + 1) − 2) =

$\dfrac{k+1}{2}(3(k+1)−1)$ is true.

Observe that

LHS:

1 + 4 + 7 + . . . + (3(k + 1) − 2) = 1 + 4 + 7 + . . . + (3k +1)

$\dfrac{k}{2}(3k − 1) + (3k + 1)$ from (1)

$\dfrac{k\ast(3k − 1) + 2(3k + 1)}{2}$

$\dfrac{3k^2 − k + 6k + 2}{ 2}$

$\dfrac{3k^2 +5k+ 2}{ 2}$

$\dfrac{(k + 1)(3k + 2)}{ 2}$ =

$\dfrac{(k + 1)(3(k + 1) − 1)}{ 2}$ =RHS

Thus by the Principle of Math Induction

$S_n$ is true for all natural numbers n.

Using mathematical induction prove that:
$\dfrac { 1 }{ 1.2.3 } +\dfrac { 1 }{ 2.3.4 } +\dfrac { 1 }{ 3.4.5 } +.....+\dfrac { 1 }{ n\left( n+1 \right) \left( n+2 \right) } =\dfrac { n\left( n+3 \right) }{ 4\left( n+1 \right) \left( n+2 \right) } .$

$\begin{array}{l} \dfrac { 1 }{ { 1.2.3 } } +\dfrac { 1 }{ { 2.3.4 } } +\dfrac { 1 }{ { 3.4.5 } } +....+\dfrac { 1 }{ { n\left( { n+1 } \right) \left( { n+2 } \right) } } =\dfrac { { n\left( { n+3 } \right) } }{ { 4\left( { n+1 } \right) \left( { n+2 } \right) } } \\ here, \\ p\left( 1 \right) :\, \dfrac { 1 }{ { 1.2.3 } } =\dfrac { 1 }{ 6 } \\ R.H.s=\dfrac { { 1.4 } }{ { 4.2.3 } } =\dfrac { 1 }{ 6 } \\ for,\, n=k \\ p\left( k \right) \, is\, also\, true \\ \dfrac { 1 }{ { 1.2.3 } } +\dfrac { 1 }{ { 2.3.4 } } +....+\dfrac { 1 }{ { k\left( { k+1 } \right) \left( { k+2 } \right) } } =\dfrac { { k\left( { k+3 } \right) } }{ { 4\left( { k+1 } \right) \left( { k+2 } \right) } } \\ for,\, \left( { k+1 } \right) \, is\, also\, true \\ \dfrac { 1 }{ { 1.2.3 } } +\dfrac { 1 }{ { 2.3.4 } } +.....+\dfrac { 1 }{ { k\left( { k+1 } \right) \left( { k+2 } \right) } } +\dfrac { 1 }{ { \left( { k+1 } \right) \left( { k+2 } \right) \left( { k+3 } \right) } } =\dfrac { { \left( { k+1 } \right) \left( { k+4 } \right) } }{ { 4\left( { k+2 } \right) \left( { k+3 } \right) } } \\ L.H.S=\dfrac { { k\left( { k+3 } \right) } }{ { 4\left( { k+1 } \right) \left( { k+2 } \right) } } +\dfrac { 1 }{ { \left( { k+1 } \right) \left( { k+2 } \right) \left( { k+3 } \right) } } \\ =\dfrac { 1 }{ { \left( { k+1 } \right) \left( { k+2 } \right) } } \left[ { \dfrac { { k\left( { k+3 } \right) } }{ 4 } +\dfrac { 1 }{ { \left( { k+3 } \right) } } } \right] \\ =\dfrac { 1 }{ { \left( { k+1 } \right) \left( { k+2 } \right) } } \left[ { \dfrac { { { k^{ 3 } }+6{ k^{ 2 } }+9k+9 } }{ { 4\left( { k+3 } \right) } } } \right] \\ =\dfrac { 1 }{ { \left( { k+1 } \right) \left( { k+2 } \right) } } .\dfrac { { \left( { k+2 } \right) \left( { k+4 } \right) \left( { k+1 } \right) } }{ { 4\left( { k+3 } \right) } } \\ =\dfrac { { \left( { k+1 } \right) \left( { k+4 } \right) } }{ { 4\left( { k+2 } \right) \left( { k+3 } \right) } } \\ =R.H.S \\ so,\, p\left( { k+1 } \right) \, is\, true \\ p\left( n \right) \, is\, true\, \forall \, n\in N. \end{array}$

Prove that $n^{2}-n$ is divisible by 2 for every positive integer.

$\textbf{Hint: We will solve this question using divisibility}$

$\textbf{Step-1: If n in an even positive integer}$

$\text{Let}$

$n=2q$

$\text{hence,}$

$n^2-n=4q^2-2q$

$n^2-n=2q(2q-1)$

$\text{which is divisible by}$

$2.$

$\textbf{Step-2: If n is an odd positive integer}$

$\text{Let}$

$n=2q+1$

$n^2-n= (2q+1)^2-(2q+1)$

$=(2q+1)(2q+1-1)$

$=2q(2q+1)$

$\text{which is divisible by}$

$2.$

$\textbf{Final Answer: Hence,}$

$\mathbf{n^2-n}$

$\textbf{is divisible by 2 for every positive integer.}$

$1 ^ { 2 } + 2 ^ { 2 } + \ldots + n ^ { 2 } > \frac { n ^ { 3 } } { 3 } , n \in \mathbf { N }$

Given that expression,

${{1}^{2}}+{{2}^{2}}+.......{{n}^{2}}>\dfrac{{{n}^{3}}}{3},n\in N$

We know that sum of square of n natural numbers,

${{1}^{2}}+{{2}^{2}}+.......{{n}^{2}}=\sum\limits_{k=1}^{n}{{{k}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}$

Hence, $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}>\dfrac{{{n}^{3}}}{3}$

$\dfrac{\left( n+1 \right)\left( 2n+1 \right)}{2}>\dfrac{{{n}^{2}}}{1}$

$\left( n+1 \right)\left( 2n+1 \right)>{{n}^{2}}$

${{n}^{2}}+3n+1>{{n}^{2}}$

$3n+1>0$

$n>\dfrac{-1}{3}$

Hence, this is the answer.

For every +ve integer 'n' prove that $7^n-3^n$ is divisible by $4$ .

$P(\eta) : 7^{\eta} - 3^{\eta}$ is divisible by

$4$

Step I :

$\eta - 1$

$\Rightarrow 7^{\eta} - 3^{\eta} = 7' - 3' = 4$

$\Rightarrow P(1)$ is true

Step II : Assume

$P(\eta)$ is true for

$\eta = m$

$\Rightarrow 7^m - 3^m$ is divisible by

$4$ .

$\Rightarrow 7^m = 3^m + 4k ; \, K\in N$

Step III :

$7^{m + 1} - 3^{m + 1}$

$= 7.7^m - 3^{m + 1}$

$= 7. (3^m + 4k) - 3^{m + 1}$

$= 7.3^m - 3^{m + 1} + 28 k$

$= 3^m (4) + 28 k$

$= 4[3^k + 7k]$

$= 4k'$

$\Rightarrow P(m + 1)$ is true

I, II and III induction says

$P(\eta)$ is true is for all

$\eta \in N$

Use the principle of mathematical induction to prove that $a+(a+d)+(a+2d)+ .... +[a+(n-1)d]=n/2[2a+(n-1)d]$

Let n=1

Then

a+(a+d)+(a+2d)+ ... +[a+(n-1)d] = a

On the other hand,

n/2[2a+(n-1)d] = 1/2 * 2a = a

So these expresions

coincide.

Supose that we have proved the

identity for all k<n+1,

in particular, for k=n we have that

a+(a+d)+(a+2d)+ ... +[a+(n-1)d] = n/2[2a+(n-1)d]

We have to prove the identity for

k=n+1, that

a+(a+d)+(a+2d)+ ... +[a+(n-1)d] + [a+nd] = (n+1)/2 * [2a+nd]

Notice that by induction

a+(a+d)+(a+2d)+ ... +[a+(n-1)d] + [a+nd]

= n/2[2a+(n-1)d] + [a+nd]

=na + n(n-1) d/2 + a + nd

=(n+1) a + (n^2/2 - n/2 + n) d

=(n+1) a + (n^2/2 + n/2) d

=(n+1) a + (n+1) n d /2

=(n+1)/2 [ 2 a + n d]

Hence prove !

Show that $n(n+1)(2n+1)$ is multiple of $6$ for every natural number $n$ .

Given

$n(n+1)(2n+1)$

$n=1$ then,

$1(1+1)(2(1)+1)=1(2)(3)=6$ which is a multiple of 6.

$n=2$ then,

$2(2+1)(2(2)+1)=2(3)(5)=30$ which is a multiple of 6.

$n=3$ then,

$3(3+1)(2(3)+1)=3(4)(7)=84$ which is also a multiple of 6.

Hence

$n(n+1)(n+1)$ is also multiple of 6 for every natural number of n.

By the principle of Mathematical induction, prove that, for $n \geq 1.$
$1 ^ { 2 } + 2 ^ { 2 } + 3 ^ { 2 } + \cdots + n ^ { 2 } > \dfrac { n ^ { 3 } } { 3 }.$

Let

$P\left(n\right)$ be the given statement, i.e,

$P\left( n\right): 1^2+2^2+.......+n^2>\dfrac{n^3}{3},n\in N$

for

$n=1$

$\because 1^2>\dfrac{1^3}{3}$ is true.

Assure that

$P\left( k\right)$ is true i.e,

$P\left(k\right) :1^2+2^2+......+k^2>\dfrac{k^3}{3}\longrightarrow \left( 1\right)$

Now we will prove that

$P\left( k+1\right)$ is true when

$P\left( k\right)$ is true.

we have,

$1^2+2^2+......+k^2+\left( k+1\right)^2$

$=\left( 1^2+2^2+........+k^2\right)+\left( k+1\right)^2$

$>\dfrac{k^3}{3}+\left( k+1\right)^2$ ( from 1 )

$=1\left[ k^3+3\left( k^2+2k+1\right)\right]$

$=\dfrac{1}{3}\left[ k^3+3k^2+6k+3\right]$

$=\dfrac{1}{3}\left[ \left( k+1\right)^3+3k+2\right]$

$>\dfrac{1}{3}\left( k+1\right)^3.$

$\Rightarrow 1^2+2^2+3^2+..........+k^2+\left( k+1\right)^2>\dfrac{1}{3}\left( k+1\right)^3$

Hence, proved.

Prove for $n\in \mathbb{N}$ .
$1.3 + {2.3^2} + {3.3^3} + ... + n{.3^n} = \dfrac{{\left( {2n - 1} \right){3^{n + 1}} + 3}}{4}$

Given

$1.3+2.3^3+3.3^3+...…..+n.3^n=\dfrac{(2n-1)3^{n+1}+3}{4}$

For

$n=1$

L.H.S

$=1.3.^1=3$

R.H.S

$=\dfrac{(R-1-1)3.^{1+1}+3}{4}$

$=\dfrac{1.3^2+3}{4}=\dfrac{9+3}{4}=\dfrac{12}{4}=3$

$\therefore$ L.H.S=R.H.S

True for

$n=1$

Let

$n=k$

$1.3+2.3^2+3.3^3+...….+k.3^k=\dfrac{(2k-1).3^{k+1}+3}{4}$

For

$n=k+1$

$1.3+2.3^2+...……+k.3^k+(k+1).3^{k+1}$

Upto k. put above value

$=\dfrac{(2k-1).3^{k+1}+3}{4}+(k+1).3^{k+1}$

$=\dfrac{(2k-1).3^{k+1}+3+4(k+1).3^{k+1}}{4}$

$=\dfrac{3^{k+1}(2k-1+4k+4)}{4}$

$=\dfrac{3.^{k+1}(6k+3)}{4}=\dfrac{3^{k+1}.3(2k+1)+3}{4}$

$=\dfrac{3^{k+1}(2(k+1)-1)+3}{4}$

Hence true for

$n=k+1$ .

Prove the following by using principle of mathematical induction for all $n \in N$ :
$1+\dfrac { 1 }{ (1+2) } +\dfrac { 1 }{ (1+2+3) } +.......+\dfrac { 1 }{ (1+2+3+\dots n) } =\dfrac { 2n }{ (n+1) } .$

Let the given statement be

$P(n)$ , i.e.

$P(n):1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+.....+\dfrac{1}{1+2+3+....n}=\dfrac{2n}{n+1}$

For

$n=1$ , we have

$P(1): 1=\dfrac{2.1}{1+1}=\dfrac{2}{2}=1$ , which is true.

Let

$P(k)$ be true for some positive integer

$k$ i.e,

$1+\dfrac{1}{1+2}+....+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+....+k}=\dfrac{2k}{k+1}$ ..........(i)

We shall now prove that

$P(k+1)$ is true.

Consider

$1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+....+\dfrac{1}{1+2+3+....+k}+\dfrac{1}{1+2+3+....+k+(k+1)}$

$=\left( 1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+....+\dfrac{1}{1+2+3+....k}\right) +\dfrac{1}{1+2+3+....+k+(k+1)}$

$=\dfrac{2k}{k+1}+\dfrac{1}{1+2+3+...+k+(k+1)}$ [ Using (i)]

$=\dfrac{2k}{k+1}+\dfrac{1}{\left( \dfrac{(k+1)(k+1+1)}{2}\right)}$

$\left[ 1+2+3+....+n=\dfrac{n(n+1)}{2}\right]$

$=\dfrac{2k}{(k+1)}+\dfrac{2}{(k+1)(k+2)}$

$=\dfrac{2}{(k+1)}\left( k+\dfrac{1}{k+2}\right)$

$=\dfrac{2}{(k+1)}\left( \dfrac{k^2+2k+1}{k+2}\right)$

$=\dfrac{2}{(k+1)}\left[ \dfrac{(k+1)^2}{k+2}\right]$

$=\dfrac{2(k+1)}{(k+2)}$

Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.

Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.

$N$ .

Prove the following by using the principle of mathematical induction for all $n\in N:$
$1+3+{ 3 }^{ 2 }+...+{ 3 }^{ n-1 }=\dfrac { \left( { 3 }^{ n }-1 \right) }{ 2 }$

Given,

$1+3+3^2+.......+3^{n-1}=\dfrac{3^n-1}{2}$

Let

$P(n)=1+3+3^2+.......+3^{n-1}=\dfrac{3^n-1}{2}$

For

$n=1$

LHS=1

RHS=

$\dfrac{3^1-1}{2}=\dfrac{2}{2}=1$

LHS=RHS

Therefore P(n) is true for

$n=1$

Assume that P(k) is true.

$P(k)=1+3+3^2+.......+3^{k-1}=\dfrac{3^k-1}{2}$

we have to prove that

$P(k+1)$ is true.

add

$3^k$ on both sides, we get,

$1+3+3^2+.......+3^{k-1}+3^k=\dfrac{3^k-1}{2}+3^k$

$P(k+1)=\dfrac{3^k-1+2(3^k)}{2}$

$=\dfrac{3(3^k)-1}{2}$

$=\dfrac{3^{k+1}-1}{2}$

Therefore

$P(k+1)$ is true when P(k) is true.

Therefore by the principle of mathematical induction, P(n) is true for n.

Solve: $9\le 1-2x$

$9\le 1-2x$

$\Rightarrow\,2x+9\le 2x+1-2x$

$\Rightarrow\,2x+9\le 1$

$\Rightarrow \,2x+9-9\le 1-9$

$\Rightarrow\,2x\le -8$

$\Rightarrow\,x\le -4$

$\therefore\,x\in\left(-\infty,-4\right]$

Prove: $\left( {2n + 7} \right) < {\left( {n + 3} \right)^2}.$ for $n\in\mathbb{N}$ .

Let

$p(n):(2n+7) < (n+3)^{2}$

For n=1

$L.H.S = (2.1+7)=2+7=9$

$R.H.S=(1+3)^{2}=4^{2}=16$

Since 9 < 16

L.H.S < R.H.S

$\therefore p(n)$ is true for n=1

Assume p(k) is true

$(2k+7) < (k+3)^{2}$ ______ (1)

we will prove that p(k+1) is true

$R.H.S = ((k+1)+3)^{2}$

$L.H.S = (2(k+1)+7)$

L.H.S

$[2(k+1)+7]$

$=2k+2+7$

$=(2k+7)+2$

Using (1) :

$(2k+7) < (k+3)^{2}$

$< (k+3)^{2}+2$

$< k^{2}+9+6k+2$

$< k^{2}+6k+11$

$< k^{2}+6k+11+(2k+5)$

$< k^{2}+8k+16$

R.H.S

$[(k+1)+3]^{2}$

$=(k+4)^{2}$

$=k^{2}+4^{2}+2.4.k$

$=k^{2}+8k+16$

L.H.S < R.H.S

$\therefore p(k+1)$ is true whenever p(k) is true.

Using Principle of Mathematical Induction, prove that
$1.3 + 2.3 ^ { 2 } + 3.3 ^ { 3 } + \ldots . n .3 ^ { n } = \dfrac { ( 2 n - 1 ) 3 ^ { n + 1 } + 3 } { 4 } ,$ where $n \in \mathrm { N }$

Let

$P(n):1.3 + 2.3 ^ { 2 } + 3.3 ^ { 3 } + \ldots . n .3 ^ { n } = \dfrac { ( 2 n - 1 ) 3 ^ { n + 1 } + 3 } { 4 } ,$
For

$n=1,$

$LHS=1.3=3;$

$RHS$

$=\dfrac {(2-1)3^2+3}{4}=\dfrac {12}{4}=3$

$\therefore LHS=RHS$

$\therefore P(1)$ is true
Let us assume

$P(k)$ is true for some

$k\in N$
i.e.,

$1.3+2.3^2+3.3^2+...+k.3^k=\dfrac {(2k-1)3^{k+1}+3}{4}$
Adding

$(k+1)^{th}$ term

$=(k+1)3^{k+1}$ on both sides, we get

$1.3+2.3^2+3.3^3+...+k.3^k+(k+1)3^{k+1}$

$=\dfrac {(2k-1)3^{k+1}}{4}+(k+1)3^{k+1}$

$=\dfrac {(2k-1)3^{k+1}+3+4(k+1)3^{k+1}}{4}$

$=\dfrac {(2k-1+4k+4)3^{k+1}+3}{4}$

$=\dfrac {3(2k+1)3^{k+1}+3}{4}$

$=\dfrac {(2k+1)3^{k+2}+3}{4}$

$=\dfrac {[2(k+1)-1]3^{(k+1)+1}+3}{4}$
which is

$P(k+1)$
Thus,

$P(k)\Rightarrow P(k+1)$
Hence, by mathematical induction

$P(n)$ is true for all

$n\in N$ .

In triangle ABC, D and E are the points on the sides AB and AC. DE||BC, BD=5 cm and DE:BC=2:Find AD and the ratio of are of triangle ADE to the quadrilateral BCED.

$DE || BC$

$\Rightarrow \triangle ADE \sim \triangle ABC$

$\Rightarrow AD/AB = DE/BC = AE /AC$

$DE/BC = 2/3$

$\Rightarrow AD/AB = 2/3$

$\Rightarrow AD/(AD+BD)=2/3$

$\Rightarrow AD/(AD+2)=2/3$

$\Rightarrow 3AD = 2AD + 4$

$\Rightarrow AD = 4$

And

Ratio of Area of Similar triangle

$=$ (Ratio of sides of similar triangle)

$^2$

$\Rightarrow$ Area of

$\triangle$ ADE / Area of

$\triangle ABC = (2/3)^2$

$\Rightarrow$ Area of

$\triangle ADE$ / Area of

$\triangle ABC = 4/9$

$\Rightarrow$ Area of

$\triangle ABC=9*$ Area of

$\triangle ADE/4$

Area of quadrilateral bced

$=$ Area of

$\Delta ABC -$ Area of

$\Delta ADE$

$\Rightarrow$ Area of quadrilateral becd

$= 9*$ Area of

$\triangle ADE / 4 -$ Area of

$\triangle ADE$

$\Rightarrow$ Area of quadrilateral bced

$=5*$ Area of

$\triangle ADE/4$

$\Rightarrow$ area of

$\triangle ADE/$ Area of quadrilateral bced

$=4/5$

Using principle of Mathematical induction prove that:
$x ^ { 2 n } - y ^ { 2 n }$ is divisible by $x + y$ , where $n \in N$

Let

$P(n):x ^ { 2 n } - y ^ { 2 n }$ is divisible by

$x + y$

For

$n=1, P(1):x^2-y^2$ is divisible by

$x + y$ which is true.

$\therefore P(1)$ is true.

Let us assume

$P(k)$ is true for some

$k\in N$

i.e,

$x^{2k}-y^{2k}$ is divisible by

$x+y$

Let

$x^{2k}-y^{2k}=(x+y)d,$ where

$d\in N$

Consider

$x^{2(k+1)}-y^{2(k+1)}=x^{2k+2}-y^{2k+2}$

$=x^{2k}x^{2}-y^{2k}y^{2}$

$=x^{2k}x^{2}-x^{2}y^{2k}+x^{2}y^{2k}-y^{2k}y^{2}$

$=x^{2}(x^{2k}-y^{2k})+y^{2k}(x^{2}-y^{2})$

$=x^{2}(x+y)d+y^{2k}(x+y)(x-y)$

$=(x+y)[x^{2}d+y^{2k}(x-y)]$

$\therefore x^{2(k+1)}-y^{2(k+1)}$ is divisible by

$x+y$

which is

$P(k+1)$

Thus,

$P(k)\Rightarrow P(k+1)$
Hence, by mathematical induction

$P(n)$ is true for all

$n\in N$ .

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$2+6+18+...+2.3^{n-1}=(3^n-1)$ .

Given: To prove by mathematical unduction,

$2 + 6 + 18 + ...+ 2.3^{n-1} = 3^{n-1}$

$\Rightarrow 2(1+3+3^2 + ..... + 3^{n-1}) = 3^{n-1}$ .

$\therefore 1 + 3 + 3^2 + ... + 3^{n-1} = \dfrac{1}{2}[3^n - 1]$

Proof: Put

$n = 1$ ,

LHS

$= 3^o = 1$

RHS

$= \dfrac{1}{2} (3-1)= \dfrac{1}{2} \times 1$

$\because$ LHS = RHS

$\therefore P(n)$ is true for

$n=1$

Now, let

$P(k)$ be true,

$\therefore 1 + 3 + 3^2 + ... 3^{k-1} = \dfrac{1}{2}(3^k-1)$

Now

$p(k+1)$ ,

$1+3+3^2 + 3^3 + ... + 3^k = \dfrac{1}{2}(3^{k+1} -1)$

Adding

$3^k$ bothsides in eqn. (1),

$1 + 3 + 3^2 + ....+3^{k-1} + 3^k = \left(\dfrac{3^k-1}{2}\right) + 3^k$

$\Rightarrow \dfrac{3^k - 1 + 2.3^k}{2}$

$\Rightarrow \dfrac{3(3^k) - 1}{2}$

$\Rightarrow \dfrac{3^{k+1}-1}{2}$

Thus,

$1+3+3^2 + ....+ 3^k = \dfrac{3^{k+1}-1}{2}$

which is same as eqn. (2)

Thus, by mathematical induction,

$P(n)$ is true for all natural numbers.

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{2^n}=\left(1-\dfrac{1}{2^n}\right)$ .

Given that:-

$\dfrac{2}{2} + \dfrac{1}{4} + \dfrac{1}{8} + .... + \dfrac{1}{2^n} = \left(1-\dfrac{1}{2^n}\right)$

To find-

Prove using mathematical Induction.

Consider,

$P(n) = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ...... + \dfrac{1}{2^n}$

For

$n = 1$ ,

$P(1) = \dfrac{1}{2^1}=\dfrac{1}{2}$ ... LHS

$\therefore$ LHS = RHS

$1-\dfrac{1}{2^1} = 1-\dfrac{1}{2} = \dfrac{1}{2}$ .... RHS

$P(1)$ is true

For

$n=2$ ,

$P(2) = \dfrac{1}{4} + \dfrac{1}{2} = \dfrac{3}{4}\to$ LHS

$\left(1 - \dfrac{1}{2^2}\right) = \left(1-\dfrac{1}{4}\right) = \dfrac{3}{4} \to$ RHS

$\therefore P(n)$ is true.

We assume,

$P(n)$ is true for all

$n = k$ .

$\therefore P(k) = \dfrac{1}{2}+\dfrac{1}{4}+ ..... + \dfrac{1}{2^k} = \left(1-\dfrac{1}{2^k}\right)$

Now, For

$n=k+1$

$\therefore P(k+1) = \dfrac{1}{2} +\dfrac{1}{4} + \dfrac{1}{8} + ..... + \dfrac{1}{2^k} + \dfrac{1}{2^{k + 1}}$

$1 - \dfrac{1}{2^k} \left(1 - \dfrac{1}{2}\right) = 1 - \dfrac{1}{2^k} \left(\dfrac{2-1}{2}\right) = 1 - \dfrac{2-1}{2^{k+1}} = \dfrac{2^{k+1}-2+1}{2^{k+1}}=1=\frac{1}{2^{k+1}}$

$P(k+1)$ is true.
Hence, the statement holds true for every value of

$n\in N$

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$1+2+3+4+....+n=\dfrac{1}{2}n(n+1)$ .

Given:

$1+2+3+... + n = \dfrac{1}{2}n (n+1)$

To prove using mathematical induction.

Proof:

Let

$P(n) : 1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$

Put

$n = 1$

LHS

$= 1$ , RHS

$= \dfrac{1\times 2}{2} = 1$

$\because$ LHS = RHS

$\therefore P(n)$ is true for

$n=1$ .

Now, Let

$P(K)$ :

$1 + 2 + 3 + .... + k = \dfrac{k(k+1)}{2}$ ... (1)

be true.

Now, we prove

$P(k+1)$ is true,

$1 + 2 + 3 + ... + k + 1 = \dfrac{(k+1)(k+1+1)}{2}$

$1 + 2 + 3 + ... + k + 1 = \dfrac{(k+1)(k+2)}{2}$ ... (2)

In eqn. (1), Add

$(k+1)$ both sides,

$1 + 2 + 3 + ... + k + (k + 1) = \dfrac{k(k+1)}{2}+(k+1)$

$1 + 2 + 3 + .... + (k+1) = \dfrac{(k+1)(k+2)}{2}$

Which is same as (2).

$\therefore P(k+1)$ is true when

$P(R)$ is true.

Hence, proved

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$1^2+3^2+5^2+7^2+....+(2n-1)^2=\dfrac{n(2n-1)(2n+1)}{3}$

Let

$P(k)$ be true for some

$k∈N$ i.e.,

$P(k): 1^2+3^2+5^2+....+(2k-1)^2=\dfrac{k(2k-1)(2k+1)}{3}$ .

Now,

$\left\{ 1^2+3^2+5^2+...+(2k-1)^2\right\}+\left\{ 2(k+1)-1\right\}^2$

$=\dfrac{k(2k-1)(2k+1)}{3}+(2k+1)^2=\dfrac{1}{3}.\left\{k(2k-1)(2k+1)+3(2k+1)^2 \right\}$

$=\dfrac{1}{3}(2k+1)\left\{k(2k-1)+3(2k+1) \right\}=\dfrac{1}{3}(2k+1)(2k^2+5k+3)$

$=\dfrac{1}{3}(k+1)(2k+1)(2k+3)$ .

Thus

$P(k+1)$ is true whenever

$P(k)$ is true.

Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$n.$

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$1.2+2.2^2+3.2^3+....+n.2^n=(n-1)2^{n+1}+2$ .

$P(n):1.2+2.2^2+3.2^3+....+n.2^n=(n-1)2^{n+1}+2$ .
For

$n=1, LHS=1.2=2; RHS=(1-1)2^{1+1}+2=2$

$\therefore LHS=RHS$

$\therefore P(1)$ is true
Let us assume

$P(k)$ is true for some

$k\in N$
i.e.,

$1.2+2.2^2+3.2^3+...+k.2^k=(k-1)2^{k+1}+2$
Adding

$(k+1)^{th}$ term

$=(k+1)2^{k+1}$ on both sides, we get

$1.2+2.2^2+3.2^3+...+k.2^k+(k+1)2^{k+1}$

$=(k-1)2^{k+1}+2+(k+1)2^{k+1}$

$=(k-1+k+1)2^{k+1}+2=2k2^{k+1}+2$

$=k2^{k+2}+2$
which is

$P(k+1)$
Thus,

$P(k)\Rightarrow P(k+1)$
Hence, by mathematical induction

$P(n)$ is true for all

$n\in N$ .

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$2+4+6+8+...+2n=n(n+1)$ .

1859791_1708722_ans_6ce6cc3e97b849d7a9a9095f9049345d.jpeg

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{(3n+1)}$

Let

$P(n):\dfrac {1}{1.4}+\dfrac {1}{4.7}+\dfrac {1}{7.10}+....+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{(3n+1)}$
For

$n=1, LHS=\dfrac {1}{1.4}=\dfrac {1}{4}$

$RHS =\dfrac {1}{3+1}=\dfrac {1}{4}$

$\therefore LHS=RHS$

$\therefore P(1)$ is true
Lte us assume

$P(k)$ is true for some

$k\in N$
i.e,

$\dfrac {1}{1.4}+\dfrac {1}{4.7}+\dfrac {1}{7.10}+....+\dfrac{1}{(3n-2)(3k-2)}=\dfrac{n}{(3k+1)}$

$=\dfrac {k}{3k+1}$
Adding

$(k+1)^{th}$ term

$=\dfrac {1}{(2k+1)(3k+4)}$

$=\dfrac {k}{3k+1}$
Adding

$(k+1)^{th}$ term

$=\dfrac {1}{(2k+1)(3k+4)}$
on both sides, we get

$\dfrac {1}{1.4}+\dfrac {1}{4.7}+\dfrac {1}{7.10}+....+\dfrac {1}{(3k-1)(3k+1)}+\dfrac {1}{(3k+1)(3k+4)}$

$=\dfrac {k}{3k+1}+\dfrac {1}{(3k+1)(3k+4)}$

$=\dfrac {k(3k+4)+1}{(3k-1)(3k+4)}=\dfrac {3k^2+4k+1}{(3k+1)(3k+4)}$

$=\dfrac {k+1}{3(k+1)+1}$
which is

$P(k+1)$
Thus,

$P(k)\Rightarrow P(k+1)$
Hence, by mathematical induction

$P(n)$ is true for all

$n\in N$ .

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$(x^{2n}-y^{2n})$ is divisible by $(x+y)$ .

Let

$P(n)$ be the statement

$x^{2n}-y^{2n}$ is divisible by

$x+y$ .
When

$n=1$ ,

$x^2-y^2=(x+y)(x-y)$ . Hence

$P(1)$ is true.

Consider

$P(k)$ is true

$\dfrac{(x^{2k}-y^{2k})}{(x+y)}=p$ .....(i).

Then,

$\left\{ x^{2(k+1)}-y^{2(k+1)}\right\}=(x^{2k}.x^2-y^{2k}.y^2)$

$=x^{2k}.x^2-x^2y^{2k}+x^2y^{2k}-y^{2k}.y^2$

$=x^2(x^{2k}-y^{2k})+y^{2k}(x^2-y^2)$

$=x^2p(x+y)+y^{2k}(x^2-y^2)$

$=(x+y)[x^2p+y^{2k}(x-y)]$ , which is divisible by

$(x+y)$ .

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$3.2^2+3^2.2^3+3^3.2^4+...+3^n.2^{n+1}=\dfrac{12}{5}(6^n-1)$ .

Let

$P(n)$ be

$3.2^2+3^2.2^3+...+3^n.2^{n+1}=\dfrac{12}{5}(6^n-1)$
When

$n=1$
LHS

$=3\times 4 =12$
RHS

$=\frac{12}{5} \times (6-1) =5$
LHS = RHS

$P(1)$ is true.
Consider

$P(k)$ is true.

$P(k):3.2^2+3^2.2^3+...+3^k.2^{k+1}=\dfrac{12}{5}(6^k-1)$ .

Now,

$P(k+1)\implies (3.2^2+3^2.2^3+.....+3^k.2^{k+1})+3^{k+1}.2^{k+2}$

$=\left\{ \dfrac{12}{5}(6^k-1)+3^{k+1}.2^{k+1}.2\right\}=\dfrac{1}{5}.\left\{12(6^k-1)+10\times 6^{k+1} \right\}$

$=\dfrac{1}{5}.\left\{2(6^{k+1}-6)+10\times 6^{k+1} \right\}=\dfrac{12}{5}(6^{k+1}-1)$ .

$P(k+1)$ is true.
The statement holds true for all natural numbers.

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$(3^{2n+2}-8n-9)$ is divisible by $8$ .

Let

$(3^{2k+2}-8k-9)=8p$ .....(i). Then,

$\left\{ 3^{2(k+1)+2}-8(k+1)-9\right\} =\left\{ 3^{(2k+2)}.3^2-8k-17\right\}=(9.3^{(2k+2)}-72k-81)+(64k+64)$

$=9.\left\{ 3^{(2k+2)}-8k-9\right\} +64(k+1)$

$=(9\times 8p)+64(k+1)$

$=8\times [9p+8(k+1)]$ , which is divisible by

$8$ .

This proves that

$3^{(2n+2)}-8n-9$ is divisible by

$8$ for all natural values of

$n$

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$n(n+1)(n+2)$ is a multiple of $6$ .

Given that-

$n(n+1)(n+2)$

To find:-

To prove 6 is a multiple of given term.

Consider,

$p(n)=n(n+1)(n+2)$ is divisible by 6.

For

$n=1$ ..

$p(1)=1\times (1+1)\times (1+2)=1\times 2\times 3=6$ ; divisible by 6.

For

$n=2$

$p(2)= 2\times (2+1)\times (2+2)=2\times 3\times 4=24$ ; divisible by 6.

Since,

$p(1), p(2),.....p(n)$ is divisible by 6.

Now, we assume, that

$p(n)$ is true for

$n=k$

$p(n)$ will also be divisible, we can write-

$k(k+1)(k+2)=6\lambda$

Now, for

$n=k+1$

$(k+1)(k+1+1)(k+2+1)=(k+1)(k+2)(k+3)=k(k+1)(k+2)+3(k+1)(k+2)$
one of

$k+1$ or

$k+2$ must be even

$3(k+1)(k+2)$ is a multiple of 6.

$(k+1)(k+2)(k+3) = 6\lambda + 6f =6(\lambda+f)$
Therefore,

$p(k+1)$ is also true.

Given statement is true.

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$\left\{ (41)^n-(14)^n\right\}$ is divisible by $27$ .

Let

$\dfrac{(41)^k-(14)^k}{27}=p$ .... (i) Then,

$\left\{ (41)^{k+1}-(14)^{k+1}\right\}=(41)^{k+1}-(41)^k.14+(41)^k.14-(14)^{k+1}$

$=(41)^k(41-14)+14\left\{ (41)^k-(14)^k\right\} =27\times (41)^k+14\times 27p$

$=27\times [(41)^k+14]$ .

This proves that

$\left\{ (41)^n-(14)^n\right\}$ is divisible by

$27$ for all natural values of

$n$

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)......\left( 1+\dfrac{1}{n}\right)=(n+1)$ .

$P(n): \left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)......\left( 1+\dfrac{1}{n}\right)=(n+1)$ .
For

$n=1, LHS=1+\dfrac {1}{1}=2$

$RHS=1+1=2$

$\therefore LHS=RHS$

$\therefore P(1)$ is true
Let us assume

$P(k)$ is true for some

$k\in N$
i.e.,

$\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)......\left( 1+\dfrac{1}{k}\right)=k+1$ .
Multiplying

$(k+1)^{th}$ term

$=\left (1+\dfrac {1}{k+1}\right)$ on both sides, we get

$\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)......\left( 1+\dfrac{1}{k}\right) \left( 1+\dfrac{1}{k+1}\right)$ .

$=(k+1)\left[ 1=\dfrac{1}{k+1}\right)=(k+1)+1$
which is

$P(k+1)$
Thus,

$P(k)\Rightarrow P(k+1)$
Hence, by mathematical induction

$P(n)$ is true for all

$n\in N$ .

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$\left(1+\dfrac{3}{1}\right)\left(1+\dfrac{5}{4}\right)\left(1+\dfrac{7}{9}\right)......\left\{ 1+\dfrac{(2n+1)}{n^2}\right\}=(n+1)^2$ .

1847237_1708780_ans_43fed61cb48d4e3aa6cfdf80b9b011df.jpeg

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$(4^n+15n-1)$ is divisible by $9$ .

Let

$(4^k+15k-1)=9p$ .....(i). Then,

$\left\{ 4^{(k+1)}+15(k+1)-1\right\}=\left\{ 4^{(k+1)}+15k+14\right\}=(4.4^k+60k-4)-45k+18$

$=4(4^k+15k-1)-9(5k-2)=(4\times 9p)-9(5k-2)$

$=9\times (4p-5k+2)$ , which is divisible by

$9$ .

This proves that

$(4^n+15n-1)$ is divisible by

$9$ for all natural values of

$n$

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$(2^{3n}-1)$ is multiple of $7$ .

Let

$(2^{3k}-1)=7p$ ... (i). Then,

$\left\{ 2^{3(k+1)}-1\right\}=(2^{3k}.2^3-1)=(8.2^{3k}-8)+7=\left\{ 8 (2^{3k}-1)+7\right\} =(8\times 7p)+7$

$=(56p+7)=7(8p+1)$ , which is divisible by

$7$ .

Thus,

$P(k + 1)$ is true whenever

$P(k)$ is true.

So, by the principle of mathematical induction

$P(n)$ is true for all natural numbers

$n.$

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$3^n \ge 2^n$

Clearly,

$3^1 \ge 2^1$ . So, the result is true for

$n=1$ .

Let it be true for

$n=k$ . Then,

$3^k \ge 2^k$ and

$3>2\Rightarrow 3^k.3 \ge 2^k.2\Rightarrow 3^{k+1} \ge 2^{k+1}$ .

So, whenever the result is true for

$k$ , then it is also true for

$(k+1)$ .

Thus,

$P(k + 1)$ is true whenever

$P(k)$ is true.

So, by the principle of mathematical induction

$P(n)$ is true for all natural numbers

$n.$

Prove that following by using the principle of mathematical induction for all $n\in N$ :
$1+3+3^2+....+3^{n-1}=\dfrac{(3^n -1)}{2}$

Let the given statement be

$P(n)$ , i.e.

$P(n): 1+3+3^2+....+3^{n-1}=\dfrac{(3^n -1)}{2}$

For

$n=1$ , we have

$P(1): =\dfrac{(3^1 -1)}{2}=\dfrac{3-1}{2}=\dfrac{2}{2}=1$ , which is true.

Let

$P(k)$ be true for some positive integer

$k$ , i.e.

$1+3+3^2+....3^{k-1}=\dfrac{(3^k -1)}{2}$ ...... (i)

We shall now prove that

$P(k+1)$ is true.

Consider

$1+3+3^2+....+3^{k-1}+3^{(k+1)-1}$

$=(1+3+3^2+....+3^{k-1})+3^k$

$=\dfrac{(3^k -1)}{2}+3^k$ [ Using (i)]

$=\dfrac{( 3^k -1)+2.3^k}{2}$

$=\dfrac{(1+2)3^k -1}{2}$

$=\dfrac{3.3^k -1}{2}$

$=\dfrac{3^{k+1}-1}{2}$

Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.

Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.

$N$ .

Prove the following by using the principle of mathematical induction for all $n\in N$ :
$1.2+2.2^2 +3.2^2+....+n.2^2 =(n-1)2^{n+1}+2$

Let the given statement be

$P(n)$ i.e.,

$P(n):1.2+2.2^2+3.2^2+....+n2^2=(n-1)2^{n+1}+2$

For

$n=1$ , we have

$P(1):1.2=2=(1-1)2^{1+1}+2=0+2=2$ which is true

Let

$P(k)$ be true for some positive integer

$k$ , i.e.,

$1.2+2.2^{2}+3.2^{2}+....+k2^{2}=(k-1)2^{k+1}+2... (i)$

We shall now prove that

$P(k+1)$ is true

Consider

$\left\{1.2+2.2^{2}+3.2^{2}+....+k2^{2}\right\}+(k+1)2^{2}$

$=(k-1)2^{1+1}+2+(k+1)2^{2}$

$2^{k+1}\left\{(k-1)+(k+1) \right\}+2$

$=2^{k+1}2k+2$

$=k2^{(k+1)+1}+2$

Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.

Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural number i.e.,

$N$ .

Prove that $3^{2n}+7$ is a multiple of 8.

1574138_1745449_ans_199c7dec20da474892b9ced9caecce6a.png

Prove the following by using the principle of mathematical induction for all $n\in N$ :
$\dfrac {1}{1.2.3}+\dfrac {1}{2.3.4}+\dfrac {1}{3.4.5}+...+\dfrac {1}{n(n+1)(n+2)}=\dfrac {n(n+3)}{4(n+1)(n+2)}$

Let the statement be

$P(n)$ i.e.,

$P(n);\dfrac {1}{1.2.3}+\dfrac {1}{2.3.4}+\dfrac {1}{3.4.5}+...+\dfrac {1}{n(n+1)(n+2)}=\dfrac {n(n+3)}{4(n+1)(n+2)}$

For

$n=1$ we have

$P(1):\dfrac {1}{1.2.3}=\dfrac {1.(1+3)}{4(1+1)(1+2)}=\dfrac {1.4}{4.2.3}=\dfrac {1}{1.2.3}$ which is true.

Let

$P(k)$ be true for some positive integer

$k$ i.e,

$\dfrac {1}{1.2.3}+\dfrac {1}{2.3.4}+\dfrac {1}{3.4.5}+...+ \dfrac {1}{k(k+1)(k+2)}=\dfrac {k(k+3)}{4(k+1)(k+2)}...(i)$

We shall now prove that

$P(k+1)$ is true,

Consider

$\left[\dfrac {1}{1.2.3}+\dfrac {1}{2.3.4}+\dfrac {1}{3.4.5}+...+\dfrac {1}{k(k+1)(k+2)}\right] +\dfrac {1}{(k+1)(k+2)(k+3)}\\$

$=\dfrac {k(k+3)}{4(k+1)(k+2)}+\dfrac {1}{(k+1)(k+2)(k+3)}\\$

$=\dfrac {1}{4(k+1)(k+2)}\left\{ \dfrac {k(k+3)}{4}+\dfrac {1}{k+3}\right\}\\$

$=\dfrac {1}{(k+1)(k+2)}\left\{\dfrac {k(k+3)^2+4}{4(k+3)}\right\}\\$

$=\dfrac {1}{(k+1)(k+2)}\left\{\dfrac {k(k^2 +6k+9)+4}{4(k+3)}\right\}\\$

$=\dfrac {1}{(k+1)(k+2)} \left\{\dfrac {k^3+6k^2+9k+4}{4(k+3)}\right\}\\$

$=\dfrac {1}{(k+1)(k+2)} \left\{\dfrac {k^3+2k^2 +k+4k^2+8k+4}{4(k+3)}\right\}\\$

$=\dfrac {1}{(k+1)(k+2)}\left\{\dfrac {k(k^2+2k+1)+4(k^2+2k+1)}{4(k+3)} \right\}\\$

$=\dfrac {1}{(k+1)(k+2)}\left\{\dfrac {k(k+1)^2+4(k+1)^2}{4(k+3)}\right\}\\$

$=\dfrac {(k+1)^2 (k+4)}{4(k+1)(k+2)(k+3)}\\$

$=\dfrac {(k+1)\left\{(k+1)+3\right\}}{4\left\{(k+1)+1\right\} \left\{(k+1)+2\right\}}\\$

Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.

Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural number i.e.,

$N$ .

Prove the following by using the principle of mathematical induction for all $n\in N$ :
$\left(1+\dfrac{3}{1}\right)\left(1+\dfrac{5}{4}\right)\left(1+\dfrac{7}{9}\right)......\left(1+\dfrac{(2n+1)}{n^{2}}\right)=(n+1)^{2}$

Let the given statement be

$P(n)$ , i.e.,

$P(n):\left(1+\dfrac{3}{1}\right)\left(1+\dfrac{5}{4}\right)\left(1+\dfrac{7}{9}\right)...... \left(1+\dfrac{2(n+1)}{k^2}\right)=(n+1)^{2}$

For

$n=1$ , we have

$P(1):\left(1+\dfrac{3}{1}\right)=4=(1+1)^{2}=2^{2}=4$ , which is true.

Let

$P(k)$ be true for some positive integer

$k$ , i.e.

$P(k):\left(1+\dfrac{3}{1}\right)\left(1+\dfrac{5}{4}\right)\left(1+\dfrac{7}{9}\right)...... \left(1+\dfrac{2(k+1)}{k^2}\right)=(k+1)^{2}$

We shall now prove that

$P(k+1)$ is true.

$\left(1+\dfrac{3}{1}\right)\left(1+\dfrac{5}{4}\right)\left(1+\dfrac{7}{9}\right) ..... \left(1+\dfrac{(2k+1)}{k^{2}}\right)\left\{1+\dfrac{\left\{2(k+1)+1\right\}}{(k+1)^{2}}\right\}$

$=(k+1)^{2}\left(1+\dfrac{2(k+1)+1}{(k+1)^{2}}\right)$

$=(k+1)^{2}\left[\dfrac{(k+1)^{2}+2(k+1)+1}{(k+1)^{2}}\right]$

$=(k+1)^{2}+2(k+1)+1$

$=\left\{(k+1)+1\right\}^{2}$

Thus,

$P(k+1)$ is true whenever

$P(k)$ true.

Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural numbers i.e.,

$N$ .

Prove the following by using the principle of mathematical induction for all $n\in N$ :
$\dfrac {1}{2}+\dfrac {1}{4}+\dfrac {1}{8}+...+\dfrac {1}{2^n}=1-\dfrac {1}{2^n}$

Let the given statement be

$P(n)$ , i.e,

$P(n), \dfrac {1}{2}+\dfrac {1}{4}+\dfrac {1}{8}+....+\dfrac {1}{2^n}=1-\dfrac {1}{2^n}$

For

$n=1$ we have

$P(1); \dfrac {1}{2}=1-\dfrac {1}{2^n}=\dfrac {1}{2}$ , which is true.

Let

$P(k)$ be terue for some positive integer

$k$ i.e.,

$\dfrac {1}{2}+\dfrac {1}{4}+\dfrac {1}{8}+....+ \dfrac {1}{2^k}=1-\dfrac {1}{2^k}....(i)$

We shall now prove that

$P(k+1)$ is true

Consider

$\left(\dfrac {1}{2}+\dfrac {1}{4}+\dfrac {1}{8}+....+\dfrac {1}{2^k}\right)+\dfrac {1}{2^{k+1}}$

$=\left(1-\dfrac {1}{2^k}\right)+\dfrac {1}{2^{k+1}}$

$1-\dfrac {1}{2^k}\left(1-\dfrac {1}{2}\right)$

$=1-\dfrac {1}{2^k}\left( \dfrac {1}{2}\right)$

$=1-\dfrac {1}{2^{k+1}}$

Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.

Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural number i.e.,

$N$ .

Prove the following by using the principle of mathematical induction for all $n\in N$ :
$1.3+3.5+5.7+....+(2n-1)(2n+1)=\dfrac {n(4n^2 +6n-1)}{3}$

Let the given statement be

$P(n)$ , i.e.,

$P(n):1.3+3.5+5.7+.....+(2n-1)(2n+1)=\dfrac {n(4n^2+6n-1)}{3}$

For

$n=1$ , we have

$P(1):1.3=3=\dfrac {1(4.1^2 +6.1-1)}{3}=\dfrac {4+6-1}{3}=\dfrac {9}{3}=3$ , which is true.

Let

$P(k)$ be true for some positive integer

$k$ , i.e.,

$1.3+3.5+5.7+.....+(2n-1)(2n+1)=\dfrac {k(4k^2+6k-1)}{3}....(i)$

We shall now prove that

$P(k+1)$ is true

$(1.3+3.5+5.7+....+ (2k-1)(2k+1))+\left\{2(k+1)-1\right\}\left\{2(k+1)+1\right\}$

$=\dfrac {k(4k^2 +6k-1)}{3}+(2k+2-1)(2k+2+1)$

$=\dfrac {k(4k^2 +6k-1)}{3}+(2k+1)(2k+3)$

$=\dfrac {k(4k^2 +6k-1)}{3}+(4k^2+8k+3)$

$=\dfrac {k(4k^2 +6k-1)+3(4k^2+8k+3)}{3}$

$=\dfrac {4k^3+6k^2-k+12k^2-24k+9}{3}$

$=\dfrac {4k^2 +18k^2+23k+9}{3}$

$=\dfrac {4k^3+14k^2+9k+4k^2+14k+9}{3}$

$=\dfrac {k(4k^2 +14k+9)+1(4k^2+14k+9)}{3}$

$=\dfrac {(k+1)(4k^2+14k+9)}{3}$

$=\dfrac {(k+1)(4k^2 +8k+4+6k+6-1)}{3}$

$=\dfrac {(k+1)(4(k^2-2k+1)+6(k+1)-1)}{3}$

$=\dfrac {(k+1)(4(k+1)^2 +6(k+1)-1)}{3}$

Thus,

$P(k+1)$ is true Whenever

$P(k)$ is true.

Hence by the principle of mathematical induction, statement

$P(n)$ is true for all natural number i.e.,

$N$ .

Prove the following by using the principle of mathematical induction for all $n\in N$ :
$\dfrac {1}{2.5}+\dfrac {1}{5.8}+\dfrac {1}{8.11}+...+\dfrac {1}{(3n-1)(3n+2)}=\dfrac {n}{(6n+4)}$

Let the given statement be

$P(n)$ , i.e,

$P(n);\dfrac {1}{2.5}+\dfrac {1}{5.8}+\dfrac {1}{8.11}+.....+ \dfrac {1}{(3n-1)(3n+2)}=\dfrac {n}{(6n+4)}$

For

$n=1$ , we have

$P(1)=\dfrac {1}{2.5}=\dfrac {1}{10}=\dfrac {1}{6.1+4}=\dfrac {1}{10}$ which is true

Let

$P(k)$ be the for the some positive integer

$k$ , i.e.,

$\dfrac {1}{2.5}+\dfrac {1}{5.8}+\dfrac {1}{8.11}+....+ \dfrac {1}{(3k+1)(3k+2)}+\dfrac {1}{\left\{3(k+1)-1 \right\} \left\{3(k+1)+2 \right\}}$

$=\dfrac {k}{6k+4}+\dfrac {1}{(3k+3-1)(3k+3+2)}$

$=\dfrac {k}{6k+4}+\dfrac {1}{(3k+2)(3k+5)}$

$=\dfrac {k}{2(3k+2)}+\dfrac {1}{(3k+2)(3k+5)}$

$=\dfrac {1}{(3k+2)}\left(\dfrac {k}{2}+\dfrac {1}{3k+5}\right)$

$=\dfrac {1}{(3k+2)}\left(\dfrac {k(3k+5)+2}{2(3k+5)}\right)$

$=\dfrac {1}{(3k+2)}\left(\dfrac {3k^2 +5k+2}{2(3k+5)}\right)$

$=\dfrac {1}{(3k+2)}\left(\dfrac {(3k+2)(k+1)}{2(3k+5)}\right)$

$=\dfrac {(k+1)}{6k+10}$

$=\dfrac {(k+1)}{6(k+1)+4}$

Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.

Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural number i.e.,

$N$ .

Prove the following by using the principle of mathematical induction for all $n\in N$ :
$a+ar+ar^2+....+ar^{n-1}=\dfrac {a(r^n -1)}{r-1}$

Let the given statement be

$P(n)$ , i.e,

$P(n);a+ar+ar^2+....+ar^{n-1}=\dfrac {a(r^n -1)}{r-1}$

For

$n=1$ , we have

$P(1); a=\dfrac {a(r^1 -1)}{(r-1)}=a$ , which is true,

Let

$P(k)$ be the true for some positive integer

$k$ , i.e

$a+ar+ar^2+....+ar^{n-1}=\dfrac {a(r^n -1)}{r-1}...(i)$

We shall now prove that

$P(k+1)$ is true

Consider

$\left\{a+ar+ar^2 +...+ ar^{k-1}\right\}+ar^{(k+1)-1}$

$=\dfrac {a(r^k -1)}{r-1}+ar^k$

$=\dfrac {a(r^k-1)+ar^k (r-1)}{r-1}$

$=\dfrac {a(r^k-1)+ar^{k-1}-ar^k}{r-1}$

$=\dfrac {ar^k-a+ar^{k+1}-ar^k}{r-1}$

$\dfrac {ar^{k+2}-a}{r-1}$

$=\dfrac {a(r^{k+1}-1)}{r-1}$

Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.

Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural number i.e.,

$N$ .

Using the principle of mathematical induction, prove that
$1+3+5+7+.....+(2n-1)=n^2$ , for all $n \in N$

Let

$P(n):1+3+5+7+...+(2n-1)=n^2$
For

$n=1, LHS=1, RHS=1^2 =1$

$\therefore LHS=RHS$

$\therefore F(1)$ is true
Let upwards assume

$P(k)$ is true for some

$k\in N$
i.e.,

$1+3+5+....+(2K-1)=k^2$
Adding

$(k+1)^{th}$ term

$=2K+1$ , on both sides, we get

$1+3+5+....+(2K-1)(2K+1)=k^2+2k+1$

$=(k+1)^2$
which is

$P(k+1)$
Thus,

$P(K)\Rightarrow P(k+1)$
Hence, by mathematical

$P(n)$ is true for all

$n\in N$

Prove that following by using the principle of mathematical induction for all $n\in N$ :
$1+2+3+.....+n < \dfrac{1}{8}(2n+1)^{2}$

Let the given statement be

$P(n)$ , i.e.

$P(n):1+2+3+....+ n\ < \dfrac{1}{8}(2n+1)^{2}$

It can be noted that

$P(n)$ is true for

$n=1$ since

$1 < \dfrac{1}{8}(2.1+1)^{2}=\dfrac{9}{8}$

Let

$P(k)$ be true for some positive integer

$k$ , i.e.

$1+2+....+k < \dfrac{1}{8}(2k+1)^{2}$ ....... (1)

We shall now prove that

$P(k+1)$ is true whenever

$P(k)$ is true.

Consider

$(1+2+....+k)+(k+1) < \dfrac{1}{8}(2k+1)^{2}+(k+1)$

$< \dfrac{1}{8} \left\{(2k+1)^{2}+8(k+1)\right\}$

$< \dfrac{1}{8}\left\{4k^{2}+4k+1+8k+8\right\}$

$< \dfrac{1}{8}\left\{4k^{2}+12k+9\right\}$

$< \dfrac{1}{8}(2k+3)^{2}$

$< \dfrac{1}{8}\left\{(k+1)+1\right\}^{2}$

Hence,

$(1+2+3+....+k)+(k+1) < \dfrac{1}{8}(2k+1)^{2}+(k+1)$

Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.

Hence, by the principle of mathematical induction statement

$P(n)$ is true for all natural numbers i.e.

$N$ .

Prove the following by using the principle of mathematical induction for all $n\in N$ :
$\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)......\left(1+\dfrac{1}{n}\right)=(n+1)$

Let the given statement be

$P(n)$ i.e.

$P(n):\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right).......\left(1+\dfrac{1}{n}\right)=(n+1)$

For

$n=1$ we have

$P(1):\left(1+\dfrac{1}{1}\right)=2=(1+1)$ , which is true.

Let

$P(k):\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)......\left(1+\dfrac{1}{k}\right)=(k+1)$ ...... (1)

We shall now prove that

$P(k+1)$ is true.

Consider

$\left[\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)......\left(1+\dfrac{1}{k}\right)\right]\left(1+\dfrac{1}{k+1}\right)$

$=(k+1)\left(1+\dfrac{1}{k+1}\right)$

$=(k+1)\left[\dfrac{(k+1)+1}{(k+1)}\right]$

$=(k+1)+1$

Thus

$P(k+1)$ is true whenever

$P(k)$ is true.

Hence, by the principle of mathematical induction statement

$P(n)$ is true for all natural numbers i.e.

$N$ .

Using the principle of mathematical induction, prove that
$1^{2}+2^{2}+3^{2}+\ldots \ldots+n^{2}=\cfrac{n(n+1)(2 n+1)}{6}$

For

$n=1, LHS=1^2=1, RHS=\dfrac {1(1+1)(2+1)}{6}=1$

$\therefore LHS=RHS$

$\therefore P(1)$ is true.
Let us assume

$P(k)$ is true for some

$k\in R$
i.e.,

$1^{2}+2^{2}+3^{2}+\ldots \ldots+k^{2}=\cfrac{k(k+1)(2 k+1)}{6}$
Adding

$(k+1)^{th}$ term

$=(k+1)^2$ , on both sides, we get,

$1^2+2^2+3^2+....... + k^2+(k+1)^2$

$=\dfrac {k(k+1)(2+1)}{6}+(k+1)^2$

$=\dfrac {(k+1)}{6}[k(2k+1)+6(k+1)]$

$=\dfrac {(k+1)}{6}[2k^2+k+6k+6]$

$=\dfrac {(k+1)}{6}(2k^2+7k+6)$

$=\dfrac {(k+1)}{6}(k+2)(2k+3)$

$=\dfrac {(k+1)[(k+1)+1][2(k+1)+1]}{6}$
which is

$P(k+1)$
Thus,

$P(k)\Rightarrow P(k+1)$
Hence, by mathematical induction

$P(n)$ is true for all

$n\in N$

Using the principle of mathematical induction, prove that
$\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{n(n+1)}=\dfrac{n}{n+1}$ , for all $n\in N$

Let

$P(n)=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{n(n+1)}=\dfrac{n}{n+1}$
For

$n=1, LHS=\dfrac {1}{1.2}=\dfrac {1}{2}; RHS=\dfrac {1}{1+1}=\dfrac {1}{2}$

$\therefore \ LHS=RHS$

$\therefore \ P(1)$ is true
Let us assume

$P(k)$ is true for some

$k\in N$
i..e,

$\dfrac {1}{1.2}+\dfrac {1}{2.3}+\dfrac {1}{3.4}+....+\dfrac {1}{k(k+1)}=\dfrac {k}{k+1}$
Adding

$(k+1)^{th}$ term

$=\dfrac {1}{(k+1)(k+2)}$ on both sides,
we get,

$\dfrac {1}{1.2}+\dfrac {1}{2.3}+....+\dfrac {1}{k(k+1)}+\dfrac {1}{(k+1)(k+2)}$

$=\dfrac {k}{k+1}+\dfrac {1}{(k+1)(k+2)}$

$=\dfrac {k(k+2)+1}{(k+1)(k+2)}$

$=\dfrac {k^2 +2k+1}{(k+1)(k+2)}$

$=\dfrac {k+1)^2}{k+1)(k+2)}=\dfrac {k+1}{k+2}$

$=\dfrac {k+1}{k+1)+1}$
which is

$P(k+1)$
Thus,

$P(k)\Rightarrow P(k+1)$
Hence, by mathematical induction

$P(n)$ is true for all

$n\in N$

Prove the following by using the principle of mathematical induction for all $n\in N$ :
$1^{2}+3^{2}+5^{2}+.....+(2n-1)^{2}=\dfrac{n(2n-1)(2n+1)}{3}$

Let the given statement be

$P(n)$ i.e.,

$P(n): 1^{2}+3^{2}+5^{2}+.....+(2n-1)^{2}=\dfrac{n(2n-1)(2n+1)}{3}$

For

$n=1$ , we have

$P(1)=1^{2}=1=\dfrac{1(2.1-1)(2.1+1)}{3}=\dfrac{1.1.3}{3}=1$ , which is true.

Let

$P(k)=1^{2}+3^{2}+5^{2}+....+(2k-1)^{2}=\dfrac{k(2k-1)(2k+1)}{3}$ ....... (1)

We shall now prove that

$P(k+1)$ is true.

Consider

$\left\{1^{2}+3^{2}+5^{2}+......+(2k-1)^{2}\right\}+\left\{2(k+1)-1\right\}^{2}$

$=\dfrac{k(2k-1)(2k+1)}{3}+(2k+2-1)^{2}$

$=\dfrac{k(2k-1)(2k+1)}{3}+(2k+1)^{2}$

$=\dfrac{2(2k-1)(2k+1)+3(2k+1)^{2}}{3}$

$=\dfrac{(2k+1)\left\{k(2k-1)+3(2k+1)\right\}}{3}$

$=\dfrac{(2k+1)\left\{2k^{2}-k+6k+3\right\}}{3}$

$=\dfrac{(2k+1)\left\{2k^{2}+5k+3\right\}}{3}$

$=\dfrac{(2k+1)\left\{2k^{2}+2k+3k+3\right\}}{3}$

$=\dfrac{(2k+1)\left\{2k(k+1)+3(k+1)\right\}}{3}$

$=\dfrac{(2k+1)(k+1)(2k+3)}{3}$

$=\dfrac{(k+1)\left\{2(k+1)-1\right\}\left\{2(k+1)+1\right\}}{3}$

Thus

$P(k+1)$ is true whenever

$P(k)$ is true.

Hence, by the principle mathematical induction, statement

$P(n)$ is true for all natural number i.e.,

$N$

Using the principle of mathematical induction, prove that
$1 \cdot 2+2 \cdot 3+\dots+n(n+1)=\cfrac{n(n+1)(n+2)}{3}$ for all $n \in N$

For

$n=1, LHS=1.2=2, RHS=\dfrac {1(1+1)(1+2)}{3}=2$

$\therefore LHS=RHS$

$\therefore P(1)$ is true.
Let us assume

$P(k)$ is true for some

$k\in N$
i.e,

$1.2+2.3+3.4+....+k(k+1)=\dfrac {k(k+1)(k+2)}{3}$
Adding

$(k+1)^{th}$ term

$=(k+1)(k+2)$ on both sides,
we get

$1.2+2.3+3.4+....+k(k+1)+(k+1)(k+2)$

$=\dfrac {k(k+1)(k+2)}{3}+(k+1)(k+2)$

$=\dfrac {(k+1)(k+2)}{3}[k+3]$

$=\dfrac {(k+1)[(k+1)+1][(k+1)+2]}{3}$
which is

$P(k+1)$
Thus,

$P(k)\Rightarrow P(k+1)$
Hence by mathematical induction

$P(n)$ is true for all

$n\in N$

Using the principle of mathematical induction, prove that
$\text { (1) } 1+2+3+\cdots+n=\cfrac{n(n+1)}{2}$ , for all $n\in N$

Let

$P(n):1+2+3+....+n=\dfrac {n(n+1)}{2}$
For

$n=1, LHS=1, RHS=\dfrac {1(1+1)}{2}=1$

$\therefore \ LHS=RHS$

$\therefore P(1)$ is true
Let us assume

$P(K)$ is true for some

$k\in N$ .
i.e.,

$1+2+3+....+k=\dfrac {k(k+1)}{2}$
Adding

$(k+1)^{th}$ term

$=k+1$ , on both sides we get

$1+2+3+....+k+(k+1)=\dfrac {k(k+1)}{2}+(k+1)$

$=\dfrac {(k+1)}{2}(k+2)$

$=\dfrac {(k+1)[(k+1)+1]}{2}$
which is

$P(k+1)$
Thus,

$P(k)\Rightarrow P(k+1)$
Hence, by mathematical induction

$P(n)$ is true for all

$n\in N$

Prove that following by using the principle of mathematical induction for all $n\in N$ :
$\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+....+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{(3n+1)}$

Let the given statement be

$P(n)$ i.e.,

$P(n):\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+....+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{(3n+1)}$

For

$n=1$ , we have

$P(1)=\dfrac{1}{1.4}=\dfrac{1}{3.1+1}=\dfrac{1}{4}+\dfrac{1}{1.4}$ which is true.

Let

$P(k)$ be true for some positive integer

$k$ i.e.,

$P(k)=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{(3k-2)(3k-1)}=\dfrac{k}{3k+1}$ ...... (1)

We shall now prove that

$P(k+1)$ is true.

Consider

$\left\{\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+......+\dfrac{1}{(3k-2)(3k+1)}\right\}+\dfrac{1}{\left\{3(k+1)-2\right\}\left\{3(k+1)+1\right\}}\\$

$=\dfrac{k}{3k+1}+\dfrac{1}{(3k+1)(3k+4)}\\$

$=\dfrac{1}{(3k+1)}\left\{k+\dfrac{1}{(3k+4)}\right\}\\$

$=\dfrac{1}{(3k+1)}\left\{\dfrac{k(3k+4)+1}{(3k+4)}\right\}\\$

$=\dfrac{1}{(3k+1)}\left\{\dfrac{3k^{2}+4k+1}{(3k+4)}\right\}\\$

$=\dfrac{1}{(3k+1)}\left\{\dfrac{3k^{2}+3k+k+1}{(3k+4)}\right\}\\$

$=\dfrac{(3k+1)(k+1)}{(3k+1)(3k+4)}\\$

$=\dfrac{(k+1)}{3(k+1)+1}$

Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.

Hence, by the principle of mathematical induction, statement

$P(n)$ is true for all natural number i,e,.

$N$ .

Prove the following by using the principle of mathematical induction for all $n\in N$ :
$\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+....+\dfrac{1}{(2n+1)(2n+3)}=\dfrac{n}{3(2n+3)}$

Let the given statement be

$P(n)$ i.e.,

$P(n):\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+.....+\dfrac{1}{(2n+1)(2n+3)}=\dfrac{n}{3(2n+3)}$

For

$n=1$ , we have

$P(1):\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+....+\dfrac{1}{(2k+1)(2k+3)}=\dfrac{k}{3(2k+3)}$ .... (1)

We shall now prove that

$P(k+1)$ is true. Consider

$\left[\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+.....+\dfrac{1}{(2k+1)(2k+3)}\right]+\dfrac{1}{\left\{2(k+1)+1\right\}\left\{2(k+1)+3\right\}}\\$

$=\dfrac{1}{3(2k+3)}+\dfrac{1}{(2k+3)(2k+5)}\\$

$=\dfrac{1}{(2k+3)}\left[\dfrac{k}{3}+\dfrac{1}{(2k+5)}\right]\\$

$=\dfrac{1}{(2k+3)}\left[\dfrac{k(2k+5)+3}{3(2k+5)}\right]\\$

$=\dfrac{1}{(2k+3)}\left[\dfrac{2k^{2}+5k+3}{3(2k+5)}\right]\\$

$=\dfrac{1}{(2k+3)}\left[\dfrac{2k^{2}+2k+3k+3}{3(2k+5)}\right]\\$

$=\dfrac{1}{(2k+3)}\left[\dfrac{2k(k+1)+3(k+1)}{3(2k+5)}\right]\\$

$=\dfrac{(k+1)(2k+3)}{3(2k+3)(2k+5)}\\$

$=\dfrac{(k+1)}{3\left\{2(k+1)+3\right\}}\\$

Thus,

$P(k+1)$ is true whenever

$P(k)$ is true.

Hence, by the principle of mathematical induction statement

$P(n)$ is true for all natural numbers, i.e.

$N$

Prove the following by using the principle of mathematical inductions for all $n\epsilon N$ $1 + 3 + 3^{2} + ... + 3^{n - 1} = \dfrac {(3^{n} - 1)}{2}$ $.

Let

$P(n):1+3+3^2+....3^{n-1}=\dfrac{3^n-1}{2}$
For

$n=1, LHS=1, RHS=\dfrac {3-1}{2}=1$

$\therefore LHS=RHS$

$\therefore P(1)$ is true
Let us assume

$P(k)$ is true for some

$k\in N$
i.e,

$1+3+3^2+...+3^{k-1}=\dfrac {3^k -1}{2}$
Adding

$(k+1)^{th}$ term

$=3^k$ on both sides we get

$1+3+3^2+...+3^{k-1}+3^k=\dfrac {3^k-1}{2}+3^k=\dfrac {3^k-1+2(3^k)}{2}=\dfrac {3(3^k)-1}{2}$

$=\dfrac {3^{k-1}-1}{2}$
which is

$P(k+1)$
Thus,

$P(k)\Rightarrow P(k+1)$
Hence, by mathematical induction

$P(n)$ is true for all

$n\in N$ .

Using principle of mathematical induction, prove that
$\cos \alpha \cos 2 \alpha \cos 4\alpha ..... \cos(2^{n-1}\alpha) = \dfrac{\sin 2^n \alpha}{2^n \sin \alpha}$ for all $n \in N$ .

Let

$P(n)$ be the statement given by

$P(n):\cos \alpha \cos 2\alpha \cos 4\alpha ......\cos (2^{n-1}\alpha) = \dfrac{\sin (2^n \alpha)}{2^n \sin \alpha}$

Setp I:

$P(1): \cos \alpha = \dfrac{\sin (2^1 \alpha)}{2^1 \sin \alpha}$

$\because \dfrac{\sin (2^1 \alpha)}{2^1 \sin \alpha} = \dfrac{\sin 2\alpha}{2\sin \alpha} = \dfrac{2\sin \alpha \cos \alpha}{2\sin \alpha} = \cos \alpha$

$\therefore P(1)$ is true.

Step II: Let

$p(m)$ be true. Then,

$\cos \alpha \cos 2\alpha \cos 4\alpha...... \cos (2^{m-1} \alpha) = \dfrac{\sin (2^m \alpha)}{2^m \sin \alpha}$

We shall now show that

$P(m+1)$ is true. For this we have to show that

$\cos \alpha \cos 2\alpha \cos 2^2 \alpha ....\cos (2^{m-1} \alpha)\cos (2^m\alpha) = \dfrac{\sin (2^{m+1}\alpha)}{2^{m+1}\sin \alpha}$

We have,

$\cos \alpha \cos 2\alpha \cos 2^2 \alpha .... \cos (2^{m-1}\alpha)\cos (2^m \alpha)$

$=\{\cos \alpha \cos 2\alpha \cos 2^2 \alpha ....\cos (2^{m-1}\alpha)\} \cos (2^m\alpha)$

$= \dfrac{\sin (2^m \alpha)}{2^m\sin \alpha}\times \cos (2^m \alpha)$

$= \dfrac{2\sin (2^m\alpha)\cos (2^m \alpha)}{2^{m+1}\sin \alpha} = \dfrac{\sin (2.2^m\alpha)}{2^{m+1}\sin \alpha} = \dfrac{\sin (2^{m+1}\alpha)}{2^{m+1}\sin \alpha}$

$\therefore P(m+1)$ is true.

Thus,

$P(m)$ is true

$\Rightarrow P(m+1)$ is true.

Hence, by the principle of mathematical induction

$P(n)$ is true for all

$n \in N$ .

Using the principle of mathematical induction, prove that
$(ab)^n=a^nb^n$ for all $n\in N$

Let

$P(n): (ab)^n=a^nb^n$
For

$n=1, LHS=(ab)^1=ab$

$RHS=a^1b^1=ab$

$\therefore LHS=RHS$

$\therefore P(1)$ is true.
Let us assume

$P(k)$ is true for some

$k\in N$ .
i.e.,

$(ab)^k=a^kb^k$
Consider

$(ab)^{k+1}=(ab)^k(ab)$

$=(a^k b^k)(ab)$

$=(a^k . a^1) (b^k . b)$

$=a^{k+1} b^{k+1}$

Using the principle of mathematical induction, prove that
$7^{2n}-3^{2n}$ is divisible by $4$ .

Let

$P(n):7^{2n}-3^{2n}$
For

$n=1, P(1):7^1-3^1=4$ which is divisible by

$4$ .

$\therefore P(1)$ is true.
Let us assume

$P(k)$ is true for some

$k\in N$
i.e.,

$7^{2n}-2^{2n}$ is divisible by

$4$
Thus,

$P(k)\Rightarrow P(k+1)$
Hence, by mathematical induction

$P(n)$ is true for all

$n\in N$

Using the principle of mathematical induction, prove that
$(1+x)^n \ge (1+nx)$ for all natural numbers $n$ , where $x > -I$ .

Let

$P(n): (1+x)^n\ge (1+nx)$ , for

$x > -1$
For

$n=1; P(1): (1+x)^1 \ge (1+x)$ which is true.
Let us assume

$P(k)$ is true for some

$k \in N$
i.e,

$(1+x)^k \ge (1+kx)$
Consider

$(1+x)^{k+1}=(1+x)^n (1+x)$

$\ge (1+kx)(1+x)$

$=(1+kx+x+kx^2)$

$\ge (1+(k+1)x)$

$\because kx^2 \ge 0$ .
which is

$P(k+1)$
Thus,

$P(k) \Rightarrow P(k+1)$
Hence, by mathematical induction,

$P(n)$ is true for all

$n\in N$ .

let O < A < $\pi$ for i = 1 , 2 , ...... n we mathematical induction to prove that
$sinA _1 \, + \,sinA _2 \, \, + \, ....... \, + \, sinA _n \,$
$\geq \, n \, sin \, \left (\dfrac{A_1 \, + \, A_2 \, + \, A_n}{n} \right )$
where n > 1 is a merital number
p sin x + (1 - p) sin p $\geq$ sin { px + ( 1 - p) y }
Where $9 \, \geq p \, \geq \, 1$ and $0 \geq x , \, y \, \geq \pi$

1969725_1003195_ans_5af237c4bb184a35a9e959c80d249cfa.jpeg

Prove that $n^3 \, + \, 3n^2 \,+\, 5n \, + \, 3$ is divisible by $3$ for any natural $n$ .

$P\left( n \right) :{ n }^{ 3 }+{ 3n }^{ 2 }+5n+3$ is divisible by 3

$\therefore P\left( 1 \right) :{ 1 }^{ 3 }+{ 3\times 1 }^{ 2 }+5\times 1+3$

$=1+3+5+3=12$ which is divisible by 3

$\therefore P\left( 1 \right)$ is true

Let

$P\left( m \right) :{ m }^{ 3 }+{ 3m }^{ 2 }+5m+3$ is divisible by 3 is true

Then, we have to prove that

$p(m+1)$ is also true

$P\left( m+1 \right) :{ \left( m+1 \right) }^{ 3 }+{ 3\left( m+1 \right) }^{ 2 }+5\left( m+1 \right) +3$

$={ m }^{ 3 }+1+3m\left( m+1 \right) +3\left( m+1+2m \right) +5m+5+3$

$={ m }^{ 3 }+3{ m }^{ 3 }+3m+3{ m }^{ 3 }+6m+5m+12$

$={ m }^{ 3 }+3{ m }^{ 3 }+5m+3+3{ m }^{ 3 }+9m+9$

$={ m }^{ 3 }+3{ m }^{ 3 }+5m+3+3\left( { m }^{ 3 }+3m+3 \right)$

$=p\left( m \right) +3\left( { m }^{ 3 }+3m+3 \right)$ which is diviesible be 3

Using the mathematical induction prove that
$tan^{-1} (1/3) \, + \,tan^{-1} (1/7) \, + \,tan^{-1} [1 / (n^2 \, + \, n \, + \, 10)]$
$= tan^{-1} [n / (\, n \, + \, 2)]$

1882079_1003203_ans_4c3fdbfb16264c41a683c04c97a5bc0f.jpeg

Show that $49^{n}+16n-1$ is divisible by $64$ for all positive integers n ?

${ (49) }^{ n }+16n−1$

$={ (1+48) }^{ n }+16n−1$

$=1+48n+.....{ 48 }^{ n }+16n−1$

$=64n+n{ C }_{ 2 }{ (48) }^{ 2 }+n{ C }_{ 3 }{ (48) }^{ 3 }+......+{ (48) }^{ n }$

$=64(n+n{ C }_{ 2 }{ (6) }^{ 2 }+n{ C }_{ 3 }{ (6) }^{ 3 }48+......+{ (6) }^{ n }{ (8) }^{ n-2 }$

$∴49n+16n−1$ is divisible by

$64$

Given $(1 + x)^{n} \geq (1 + nx)$ , for all natural number n, where $x > -n$ . What will be the value of $n$

Let

$P(n)$ be the mathematical statement given by

$P(n) : (1+x)^{n} \ge (1+nx)$ , when

$n \in N$ and

$x > -n$

Let

$n=1$

Then

$P(1) : (1+x) \ge (1+x)$ which is true.

$\therefore P(1)$ is true.

Let us consider

$P(m)$ is true.

$\therefore (1+x)^{m} \ge (1+ mx)$ ........

$(1)$

$\because x > - n \Rightarrow x+ n > 0$ .

Now, if we take

$n=1. \Rightarrow x+1 > 0$ .

Multiplying both sides of

$(1)$ by

$(x+1)$

$(1+x)^{m+1} \ge (1+x) (1+mx)$

or,

$(1+x)^{m+1} \ge 1+ (m+1) x+ mx^{2}$

or,

$(1+x)^{m+1} \ge 1+ (m+1)_{x} [\because mx^{2} \ge 0]$ .........

$(2)$

$\therefore P(n)$ is true for all values of

$n e N$ and

$x > -1$

So, the value of

$n=1$

$1+2+3+..............n<\dfrac{1}{8}(2n+1)^2$

TO PROVE :

$1+2+3+....n<\dfrac { 1 }{ 8 } { \left( 2n+1 \right) }^{ 2 }$

PROOF:

$P\left( n \right) :1+2+3+....n<\dfrac { 1 }{ 8 } { \left( 2n+1 \right) }^{ 2 }$

$P\left( 1 \right) :1<\dfrac { 1 }{ 8 } { \left( 2+1 \right) }^{ 2 }$

$\Rightarrow 1<\dfrac { 1 }{ 8 } \times 9$

$\Rightarrow 1<\dfrac { 9 }{ 8 }$ which is true.

$\therefore P(1)$ is true.

Let

$P(m)$ is true.

Then,

$P\left( m \right) :1+2+3+....m<\dfrac { 1 }{ 8 } { \left( 2m+1 \right) }^{ 2 }$

Now, we prove it for

$P(m+1)$

Then,

$P\left( m \right) :1+2+3+....m+\left( m+1 \right)$

$<\dfrac { 1 }{ 8 } { \left( 2m+1 \right) }^{ 2 }+\left( m+1 \right)$

$<\dfrac { 1 }{ 8 } { \left( { 4m }^{ 2 }+1+4m \right) }+\left( m+1 \right)$

$<\dfrac { { 4m }^{ 2 }+1+4m+8m+8 }{ 8 }$

$<\dfrac { { 4m }^{ 2 }+12m+9 }{ 8 }$

$<\dfrac { 1 }{ 8 } { \left( 2m+3 \right) }^{ 2 }$

$<\dfrac { 1 }{ 8 } { \left[ 2\left( m+1 \right) +1 \right] }^{ 2 }$ (Proved)

$\therefore P(m+1)$ is also true (proved).

Using induction method of otherwise, prove that for any non-negative integers m,n,r and k
$\sum_{m \, = \, 0}^{k}(n \, - \, m) \, \dfrac{(r \, + \, m)!}{m!} \, = \, \dfrac{(r \, + \, k \, + \, 1)!}{k!}$ $\left[\dfrac{n}{r \, + \, 1} \, - \, \dfrac{k}{r \, + \, 2}\right]$

1970761_1003230_ans_fecbec1fbbc5443680ef1023cf992fa0.jpeg

Use Euclid's division lemma to show that any positive odd integer is of the form $6q+1$ , or $6q+3,\ 6q+5$ where $q$ is some integers.

According to Euclid's division lemma, there exist integers

$a, b, q$ and

$r$ such that

$a = bq + r$ , where

$0 \le r < b$

Now let us assume,

$b = 6$

Then we have,

$a=6q+r$ , where

$0 \le r < 6$

thus,

$r$ can take values

$0, 1, 2, 3, 4, 5$

Consider the equation,

$a = 6q+r$

Case 1: When

$r = 0$

Thus,

$a = 6q$

Rewriting the above equation, we have

$a = 2(3q)$

Hence

$a$ is an even number

Case 2: When

$r=1$

Thus,

$a=6q+1$

Rewriting the above equation, we have

$a = 2\times 3q+1$

$=2m+1$ , where

$m = 3q$

Hence

$a$ is an odd number

Case 3: When

$r=2$

Thus,

$a=6q+2$

Rewriting the above equation, we have

$a = 2\times (3q+1)$

$=2m$ , where

$m = 3q+1$

Hence

$a$ is an even number

Case 4: When

$r=3$

Thus,

$a=6q+3$

Rewriting the above equation, we have

$a = 2\times (3q+1)+1$

$=2m+1$ , where

$m = 3q+1$

Hence

$a$ is an odd number

Case 5: When

$r=4$

Thus,

$a=6q+4$

Rewriting the above equation, we have

$a = 2\times (3q+2)$

$=2m$ , where

$m = 3q+2$

Hence

$a$ is an even number

Case 6: When

$r=5$

Thus,

$a=6q+5$

Rewriting the above equation, we have

$a = 2\times (3q+2)+1$

$=2m+1$ , where

$m = 3q+2$

Hence

$a$ is an odd number

Therefore, any positive odd integer is of the form

$6m, 6m+1$ or

$6m+5$ , where

$m$ is some integer.

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{(3n-1)(3n+2)}=\dfrac{n}{(6n+4)}$ .

1948206_1708777_ans_197b1e3528af401ebc6aad5123e49827.jpg

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{(2n-1)(2n+1)}=\dfrac{n}{(2n+1)}$

1948205_1708774_ans_027241eb387043028875042a97c0249c.jpg

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$1+3+3^2+3^3+....+3^{n-1}=\dfrac{1}{2}(3^n-1)$ .

1861441_1708728_ans_8c17d8da632f4aa89dc5e20d8ca4820e.jpeg

If $A = \left[ \begin{array} { c c } { \cos \theta } & { \sin \theta } \\ { - \sin \theta } & { \cos \theta } \end{array} \right]$ then show that for all the positive integers $n$ $A ^ { n } = \left[ \begin{array} { c c } { \cos n \theta } & { \sin n \theta } \\ { - \sin n \theta } & { \cos n \theta } \end{array} \right]$

$A=\left(\begin{matrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{matrix}\right)$

$A^2=\left(\begin{matrix} \cos\theta & \sin\theta\\ -\sin\theta &\cos\theta\end{matrix}\right)\left(\begin{matrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{matrix}\right)$

$=\left(\begin{matrix} \cos^2\theta -\sin^2\theta & \sin\theta\cos\theta +\sin\theta\cos\theta\\ -\cos\theta\sin\theta -\cos\theta\sin\theta & \cos^2\theta -\sin^2\theta\end{matrix}\right)$

$=\left(\begin{matrix} \cos 2\theta & \sin 2\theta\\ -\sin 2\theta & \cos 2\theta\end{matrix}\right)$

The result is true for

$n=1$ and

$2$

Let us assume the result is true for

$n=k$

i.e.,

$A^k=\left(\begin{matrix} \cos k\theta & \sin k\theta\\ -\sin k\theta & \cos k\theta\end{matrix}\right)$

We have to prove the result for

$n=k+1$

$A^{k+1}=A^k\cdot A$

$=\left(\begin{matrix} \cos k\theta & \sin k\theta\\ -\sin k\theta & \cos k \theta\end{matrix}\right)\left(\begin{matrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{matrix}\right)$

$=\left(\begin{matrix} \cos k\theta \cos\theta -\sin k\theta \sin\theta & \cos k\theta \sin\theta +\sin k\theta \cos\theta\\ -(\sin k\theta \cos \theta +\cos k\theta \sin\theta) & \cos k\theta \cos\theta -\sin k\theta \sin\theta \end{matrix}\right)$

$=\left(\begin{matrix} \cos (k+1)\theta & \sin(k+1)\theta\\ -\sin(k+1)\theta & \cos(k+1)\theta\end{matrix}\right)$

The result is true for

$n=k+1$ also. Hence by the principle of Mathematical Induction, we can understand that the result is true for all

$n\in N$ .

Show that $n(n^2-1)(3n+2)$ is divisible by 24.

1948364_1745446_ans_de6fc1531a8e404a88a1689108125102.jpg

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$(x^{2n}-1)$ is divisible by $(x-1)$ , where $x\neq 1$ .

The expression is given as

${ x }^{ 2n }-1$

Let us assume that this expression is divisible by

$\left( x-1 \right)$

In other words,

$\left( x-1 \right)$ is one of the factors of

${ x }^{ 2n }-1$

$\therefore { x }^{ 2n }-1=\left( x-1 \right) P\left( x \right)$

where,

$P\left( x \right)$ = polynomial in x

Now, Put n = k in above equation.

$\therefore { x }^{ 2k }-1=\left( x-1 \right) P\left( x \right)$ Equation (1)

Now, put n = k+1 in above equation,

$\therefore LHS={ x }^{ 2(k+1) }-1$

$={ x }^{ 2k }.{ x }^{ 2 }-1$

$={ x }^{ 2k }.{ x }^{ 2 }-{ x }^{ 2 }+{ x }^{ 2 }-1$

$=\left( { x }^{ 2k }.{ x }^{ 2 }-{ x }^{ 2 } \right) +\left( { x }^{ 2 }-1 \right)$

$={ x }^{ 2 }\left( { x }^{ 2k }-1 \right) +\left( { x }+1 \right) \left( { x }-1 \right)$

From equation (1), we can write,

$LHS={ x }^{ 2 }\left( { x }-1 \right) P\left( x \right) +\left( { x }+1 \right) \left( { x }-1 \right)$

$=\left( { x }-1 \right) \left[ { x }^{ 2 }P\left( x \right) +\left( { x }+1 \right) \right]$

Let

$Q\left( x \right) ={ x }^{ 2 }P\left( x \right) +\left( { x }+1 \right)$

$\therefore LHS=\left( { x }-1 \right) Q\left( x \right)$

Thus, we can conclude that for n = k+1 also, expression is divisible by

$\left( { x }-1 \right)$

Thus, by principle of mathematical induction, we can say that,

${ x }^{ 2n }-1$ is divisible by

${ x }-1$ for all

$n\in N$ where

$x\neq 1$ (For x = 1, expression will become zero)

Using the principle of mathematical induction, prove the following for all $n\in N$ :
$1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+....+\dfrac{1}{(1+2+3+....+n)}=\dfrac{2n}{(n+1)}$ .

1847199_1708765_ans_ab3f202ee1a34b00ad92d1841820df42.jpg

Prove that $^n$ C $_r$ + $^{n - 1}$ C $_r$ + $^{n - 2}$ C $_r$ + $^{n - 2}$ C $_r$ +....................+ $^r$ C $_r$ = $^{n +1}$ C $_{r + 1}$

To prove :

${ n }_{ { C }_{ r } }+{ n-1 }_{ { C }_{ r } }+{ n-2 }_{ { C }_{ r } }+......+{ n }_{ { C }_{ r } }={ n+1 }_{ { C }_{ r+1 } }$

We will prove this by induction

${ S }_{ n }={ r }_{ { C }_{ r } }+{ r+1 }_{ { C }_{ r } }+{ r+2 }_{ { C }_{ r } }+......+{ n-2 }_{ { C }_{ r } }+{ n-1 }_{ { C }_{ r } }+{ n }_{ { C }_{ r } }\quad \quad ,\quad r\le n$

$=\sum _{ K=0 }^{ n-r }{ \quad } { r+K }_{ { C }_{ r } }={ n+1 }_{ { C }_{ r+1 } }$ or

${ n+1 }_{ { C }_{ n-r } }\quad ,\quad r\le n$

We are tying to prove the above statement

Now,

${ S }_{ n=r }={ S }_{ r }={ n }_{ { C }_{ r } }=1$

and,

RHS

$={ r+1 }_{ { C }_{ r+1 } }=1$ as

$n=r$

Also,

${ S }_{ r+1 }={ r+1 }_{ { C }_{ r } }+{ r+1 }_{ { C }_{ r } }=1+r+1=r+2;$ RHS with

$n=r+1$ is equal to

${ r+1+1 }_{ { C }_{ r+1 } }=r+2$

Say

${ S }_{ P }={ P+1 }_{ { C }_{ r+1 } }$

Then

${ S }_{ PH }={ P+1 }_{ { C }_{ r+1 } }+{ P+1 }_{ { C }_{ r } }=\dfrac { \left( PH \right) ! }{ \left( r+1 \right) !\left( p-r \right) ! } +\dfrac { \left( P+1 \right) ! }{ r!\left( p-r+1 \right) ! }$

$=\dfrac { \left( P+1 \right) ! }{ r!\left( p-r \right) ! } \left[ \dfrac { 1 }{ r+1 } +\dfrac { 1 }{ \left( PH \right) -r } \right] =\dfrac { \left( P+1 \right) ! }{ r!\left( p-r \right) ! } \times \dfrac { P+2 }{ \left( r+1 \right) \left( pH-r \right) }$

$=\dfrac { \left( P+2 \right) ! }{ \left( r+1 \right) !\left( p+1-r \right) ! } ={ P+2 }_{ { C }_{ r+1 } }=RHS$

So it

${ S }_{ p }$ is true the

${ S }_{ p+1 }$ is true for all

$P$ .

Hence proved.

Principle Of Mathematical Induction - Class 11 Engineering Maths - Extra Questions

Prove that 2n>n2^n>n for all positive integers nn

For any natural number n , prove that inequality |sin n x| ≥\geq n | sin x |

Prove that 7 is a factor of 23n2^{3n} - 1 for all natural numbers n.

Prove the assertions of the following problems Prove that 2^n > n^22^n > n^2 for any natural number n > 5.n > 5.

If P(n)P(n) is the statement ''2^n\ge 3n"''2^n\ge 3n" and if P(r)P(r) is true prove that P(r+1)P(r+1) is true.

Using the principle of mathematical induction prove that 2 + 4 + 6 + .... + 2n = {n^2} + n2 + 4 + 6 + .... + 2n = {n^2} + n.

Prove by mathematical induction,{ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+....+{ n }^{ 2 }=\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+....+{ n }^{ 2 }=\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 }

Using the principle of mathematical induction, show that;2+4+6+.....+2n=n^2+n2+4+6+.....+2n=n^2+n

Using Principle of mathematical induction prove that 6^n -16^n -1 divisible by 55.

Using the principle of Mathematical Induction, prove the following for all nn and NN1) {3}^{n}>{2}^{n}{3}^{n}>{2}^{n}

Prove that the square of any positive integer of the form 5m+1 will leave a remainder 1 when divided by 5 for some integer m

If P(n)P(n) is the statement n^{2}-n+41n^{2}-n+41 is prime, prove that P(1),P(2)P(1),P(2) and P(3)P(3) are true.Prove also that P(41)P(41) is not true.

Solve: (2n+7)<(n+3)^2(2n+7)<(n+3)^2

Using P.M.I. prove that{n^2} < {2^n}\forall n \ge 5{n^2} < {2^n}\forall n \ge 5

Use the principle of mathematical induction to show that : 2 + 2^2 + ..... + 2^n = 2^{n+1} - 2 2 + 2^2 + ..... + 2^n = 2^{n+1} - 2 for every natural number n.n.

Show that only one of the no n, n + 2n, n + 2 and n + 4n + 4 in divisible 33.

Prove by method of induction, for all n \in n \in 1^2 + 2^2 + 3^2 + ... + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}1^2 + 2^2 + 3^2 + ... + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}.

To prove: n \le 2^{n}n \le 2^{n} for n\ in\ Nn\ in\ NP(n):n \le 2^{n}.........P(n):n \le 2^{n}......... for n\ in\ Nn\ in\ N

Can the number 6^{n}6^{n}, where n is a natural number, end with digit 5? Give reasons.

For every natural number n(>1)n(>1), prove that 4^{n}+15n-14^{n}+15n-1 is divisible by 99.

Prove by method of induction; for all n \in Nn \in N 3 + 7 + 11 + \ldots . . . +3 + 7 + 11 + \ldots . . . + to nn terms = n ( 2 n + 1 )= n ( 2 n + 1 )

Prove that 3^{n+1} > 3(n+1)3^{n+1} > 3(n+1).

"The product of three consecutive positive integers is divisible by 6". Is.this statement true or false"? Justify your answer.

Prove that 1 + 2 + 3 + ..... + n =1 + 2 + 3 + ..... + n = \displaystyle \frac {n(n\, +\, 1)}{2}\displaystyle \frac {n(n\, +\, 1)}{2}.

Prove that 3^{n + 1} > 3(n + 1)3^{n + 1} > 3(n + 1).

Prove 1+3+5+\cdots+(2n-1)=n^21+3+5+\cdots+(2n-1)=n^2

Prove 3^{n+1}>3^{n+1}>3(n+1)3(n+1)

Prove the following using the principle of mathematical induction for all n\in Nn\in N:1+3+{3}^{2}+........+{3}^{n-1}=\cfrac{({3}^{n}-1)}{2}1+3+{3}^{2}+........+{3}^{n-1}=\cfrac{({3}^{n}-1)}{2}

Prove that 1 + 3 + 5 + ..... + (2n - 1) = n1 + 3 + 5 + ..... + (2n - 1) = n ^2^2.

Prove the following by using principle of mathematical induction for all n\in N: 1.3+3.5+5.7+.......+(2n-1)(2n+1)=\cfrac{n(4{n}^{2}+6n-1)}{3}n\in N: 1.3+3.5+5.7+.......+(2n-1)(2n+1)=\cfrac{n(4{n}^{2}+6n-1)}{3}

Prove the following by using the principle of mathematical induction for all n\in N: 1.3+2.{3}^{2}+3.{3}^{3}+.....+n.{3}{^n}=\cfrac{(2n-1){3}^{n+1}+3}{4}n\in N: 1.3+2.{3}^{2}+3.{3}^{3}+.....+n.{3}{^n}=\cfrac{(2n-1){3}^{n+1}+3}{4}

Prove the following by using the principle of mathematical induction for all n\in N:1.2.3+2.3.4+......+n(n+1)(n+2)=\cfrac{n(n+1)(n+2)(n+3)}{4}n\in N:1.2.3+2.3.4+......+n(n+1)(n+2)=\cfrac{n(n+1)(n+2)(n+3)}{4}

Prove the following by using the principle of mathematical induction for all n \in N: \cfrac{1}{2}+\cfrac{1}{4}+\cfrac{1}{8}+.....+\cfrac{1}{{2}^{n}}=1-\cfrac{1}{{2}^{n}}n \in N: \cfrac{1}{2}+\cfrac{1}{4}+\cfrac{1}{8}+.....+\cfrac{1}{{2}^{n}}=1-\cfrac{1}{{2}^{n}}

Prove the following by using the principle of mathematical induction for all n\in n\in :1+\cfrac{1}{(1+2)}+\cfrac{1}{(1+2+3)}+..........+\cfrac{1}{1+2+3+.....n)}=\cfrac{2n}{(n+1)}1+\cfrac{1}{(1+2)}+\cfrac{1}{(1+2+3)}+..........+\cfrac{1}{1+2+3+.....n)}=\cfrac{2n}{(n+1)}

Prove the following by using the principle of mathematical induction for all n\in N: 1.2+2.3+3.4+......+n(n+1)=\left[ \cfrac { n(n+1)(n+2) }{ 3 } \right] n\in N: 1.2+2.3+3.4+......+n(n+1)=\left[ \cfrac { n(n+1)(n+2) }{ 3 } \right]

Prove the following by using the principle of mathematical induction for all n\in Nn\in N{ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+.......+{ n }^{ 3 }={ \left[ \cfrac { n(n+1) }{ 2 } \right] }^{ 2 }{ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+.......+{ n }^{ 3 }={ \left[ \cfrac { n(n+1) }{ 2 } \right] }^{ 2 }

Prove the following by using the principle of mathematical induction for all n\in N: 1.2+2.{2}^{2}+3.{2}^{2}+.......+n.{2}^{n}=(n-1){2}^{n+1}+2n\in N: 1.2+2.{2}^{2}+3.{2}^{2}+.......+n.{2}^{n}=(n-1){2}^{n+1}+2

Prove the following by using the principle of mathematical induction for all n\in N: \cfrac{1}{2.5}+\cfrac{1}{5.8}+\cfrac{1}{8.11}+......+\cfrac{1}{(3n-1)(3n+2)}=\cfrac{n}{(6n+4)}n\in N: \cfrac{1}{2.5}+\cfrac{1}{5.8}+\cfrac{1}{8.11}+......+\cfrac{1}{(3n-1)(3n+2)}=\cfrac{n}{(6n+4)}

Prove the following by using the principle of mathematical induction for all n\in N:{10}^{2n-1}+1n\in N:{10}^{2n-1}+1 is divisible by 1111

Prove the following by using the principle of mathematical induction for all n\in N:{x}^{2n}-{y}^{2n}n\in N:{x}^{2n}-{y}^{2n} is divisible by x+yx+y.

Prove the following b y using the principle of mathematical induction for all n\in N:1+2+3+.....+n<\cfrac{1}{8}{(2n+1)}^{2}n\in N:1+2+3+.....+n<\cfrac{1}{8}{(2n+1)}^{2}

Prove the following by using the principle of mathematical induction for all n\in N: P(n):a+ar+a{r}^{2}+......+a{r}^{n-1}=\cfrac{a({r}^{n}-1)}{r-1}n\in N: P(n):a+ar+a{r}^{2}+......+a{r}^{n-1}=\cfrac{a({r}^{n}-1)}{r-1}

Prove the following by using the principle of mathematical induction for all n\in N:{ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+.......+({ 2n-1) }^{ 2 }=\cfrac { n(2n-1)(2n+1) }{ 3 } n\in N:{ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+.......+({ 2n-1) }^{ 2 }=\cfrac { n(2n-1)(2n+1) }{ 3 }

Prove the following by using the principle of mathematical induction for all n\in N:n(n+1)(n+5)n\in N:n(n+1)(n+5) is multiple of 33.

Find counter examples to disprove the following statement.A quadrilateral with all sides are equal is a square.

Using mathematical induction, prove that \cfrac{d}{dx}({x}^{n})=n{x}^{n-1}\cfrac{d}{dx}({x}^{n})=n{x}^{n-1} for all positive integers nn.

Prove the following by using the principle of mathematical induction for all n\in N:{41}^{n}-{14}^{n}n\in N:{41}^{n}-{14}^{n} is a multiple of 2727.

Prove the following by using the principle of mathematical induction for all n\in N:(2n+7)< {(n+3)}^{2}n\in N:(2n+7)< {(n+3)}^{2}.

Prove the following by using the principle of mathematical induction for all n\in N:{3}^{2n+2}-8n-9n\in N:{3}^{2n+2}-8n-9 is divisible by 88.

Prove by induction:{1}^{2}+{2}^{2}+{3}^{2}+........+{n}^{2}=\cfrac{1}{6}n(n+1)(2n+1){1}^{2}+{2}^{2}+{3}^{2}+........+{n}^{2}=\cfrac{1}{6}n(n+1)(2n+1)

Prove by induction:{x}^{n}-{y}^{n}{x}^{n}-{y}^{n} is divisible by x+yx+y when nn is even.

Show that { 2 }^{ 2n }-3n-1{ 2 }^{ 2n }-3n-1 is divisible by 99 for all positive integral values of nn.

Prove by induction:2+{2}^{2}+{2}^{3}+........+{2}^{n}=2({2}^{n}-1)2+{2}^{2}+{2}^{3}+........+{2}^{n}=2({2}^{n}-1)

Prove the following by the principle of mathematical induction: (n\epsilon N)(n\epsilon N)1^{2} . 2 + 2^{2} . 3 + .... + n^{2} (n + 1) = \dfrac {n(n + 1)(n + 2)(3n + 1)}{12}1^{2} . 2 + 2^{2} . 3 + .... + n^{2} (n + 1) = \dfrac {n(n + 1)(n + 2)(3n + 1)}{12}.

Prove the following by the principle of Mathematical Induction.(2n + 1)(2n - 1)(2n + 1)(2n - 1) is an odd number for all n\epsilon Nn\epsilon N.

Prove by induction:\cfrac{1}{1.2}+\cfrac{1}{2.3}+\cfrac{1}{3.4}+.....\cfrac{1}{1.2}+\cfrac{1}{2.3}+\cfrac{1}{3.4}+..... to nn terms=\cfrac{n}{n+1}=\cfrac{n}{n+1}

Prove by induction: 1+3+5+.....+(2n-1)={n}^{2}1+3+5+.....+(2n-1)={n}^{2}

If pp and qq are any two positive integers, show that |\underline {pq}|\underline {pq} is divisible by (|\underline {p})^{q} . |\underline {q}(|\underline {p})^{q} . |\underline {q} and by (|\underline {q})^{p} . |\underline {p.}(|\underline {q})^{p} . |\underline {p.}

(1 + x)^n >1 (1 + x)^n >1 + nx for n > 2 , n \epsilon \epsilon N x > -1 x \neq \neq 0

Prove 2 + 5 + 8 + 11 + ... + (3n - 1) =2 + 5 + 8 + 11 + ... + (3n - 1) = \dfrac{1}{2}\dfrac{1}{2} n (3n + 1) n n (3n + 1) n \epsilon\epsilon N

Prove that 11^{n +2} + 12^{2n \, + \, 1}11^{n +2} + 12^{2n \, + \, 1} is divisible by 133133 for any non-negative integral nn.

Prove \dfrac {1}{1 \cdot 2\cdot 3 } \, + \, \dfrac {1}{2 \cdot 3\cdot 4 } \, + \, \dfrac {1}{n (n + 1)(n + 2)} \, = \, \dfrac{1(n + 3)}{4(n + 1)(n + 2)}\dfrac {1}{1 \cdot 2\cdot 3 } \, + \, \dfrac {1}{2 \cdot 3\cdot 4 } \, + \, \dfrac {1}{n (n + 1)(n + 2)} \, = \, \dfrac{1(n + 3)}{4(n + 1)(n + 2)}

\sqrt 2 + \sqrt 2 + \sqrt 2 + ...+ n\ terms = 2\cos \dfrac{\pi}{2^{n +1}} , n \epsilon N\sqrt 2 + \sqrt 2 + \sqrt 2 + ...+ n\ terms = 2\cos \dfrac{\pi}{2^{n +1}} , n \epsilon N

Prove the assertions of the following problems Prove that the expression n^3 \, - \, nn^3 \, - \, n is divisible by 2424 for any odd nn.

For any natural number n > 1 \dfrac {1}{n \, + \, 1} \, + \, \dfrac {1}{n \, + \, 2} \, + \, ..... \, + \, \dfrac {1}{2n} > \dfrac {13}{14} \dfrac {1}{n \, + \, 1} \, + \, \dfrac {1}{n \, + \, 2} \, + \, ..... \, + \, \dfrac {1}{2n} > \dfrac {13}{14}

Prove that induction thatsin x + sin 3x + .. + sin(2n - 1) x = \dfrac {sin^2 nx}{sin x} \dfrac {sin^2 nx}{sin x}

Observing that 1^3 \, = \, 1, \, 2^3 \, = \, 3 \, + \, 5,1^3 \, = \, 1, \, 2^3 \, = \, 3 \, + \, 5,3^33^3 = 7 + 9 + 11, 4^34^3 = 13 + 15 + 17 + 19Find a general formula for the cube of natural number n and prove it by the principle of mathematical induction

Explain the method of mathematical induction and use it to show that 11^{n \, + \, 2} \, + \, 12^{2n \, + \, 1}11^{n \, + \, 2} \, + \, 12^{2n \, + \, 1} where n is natural number is divisible by 133

If x_1x_2x_3 \, .... \, x_n \ = \, 1(x_1 \, > \, 0, i \, = \, 1,2,.....n),x_1x_2x_3 \, .... \, x_n \ = \, 1(x_1 \, > \, 0, i \, = \, 1,2,.....n), prove that x_1 \, + \, x_2 \, + \, .....x_n \, \geq \, n \, (n \, \geq \, 2)x_1 \, + \, x_2 \, + \, .....x_n \, \geq \, n \, (n \, \geq \, 2)

Prove ^{2n} C _n \, > \, \dfrac{4^n }{n + 1}^{2n} C _n \, > \, \dfrac{4^n }{n + 1}

Prove \dfrac {2n!}{(n!)^2} \, > \, \dfrac {4n}{2n \, + \, 1} \dfrac {2n!}{(n!)^2} \, > \, \dfrac {4n}{2n \, + \, 1}

If 0 < \alpha < \, \dfrac{\pi}{4(n \, - \, 1)} \alpha < \, \dfrac{\pi}{4(n \, - \, 1)} where n > 1 , then prove that \tan n\tan n \alpha \, > \, n \, \tan \, \alpha \alpha \, > \, n \, \tan \, \alpha

Show by the mathematical induction that \dfrac{1}{sin 2 x } \, + \, \dfrac{1}{sin 4 x } \, + \, \dfrac{1}{sin 2^n x } \, = \, cot x \, - \, cot 2^n \, x \dfrac{1}{sin 2 x } \, + \, \dfrac{1}{sin 4 x } \, + \, \dfrac{1}{sin 2^n x } \, = \, cot x \, - \, cot 2^n \, x

Prove that { 1 }^{ 2 }+{ 2 }^{ 2 }+...+{ n }^{ 2 }>\cfrac { { n }^{ 3 } }{ 3 } ,n\epsilon N{ 1 }^{ 2 }+{ 2 }^{ 2 }+...+{ n }^{ 2 }>\cfrac { { n }^{ 3 } }{ 3 } ,n\epsilon N

Prove the following by using the first principle 1+3 + 3^2 + ......+ 3^{n-1} =\frac{(3^n -1)}{2} 1+3 + 3^2 + ......+ 3^{n-1} =\frac{(3^n -1)}{2}.

Prove the following by PMI for all nn belong to NN1\times3+3\times5+5\times7 .........+(2n-1)(2n+1)=\dfrac{n(4n^2+6n-1)}{3}1\times3+3\times5+5\times7 .........+(2n-1)(2n+1)=\dfrac{n(4n^2+6n-1)}{3}

Prove that for any positive interger n, n^3 -nn, n^3 -n is divisible by 66.

Let p\geq\geq 3 be an integer and \alpha\alpha , \beta\beta be the roots of x^2x^2 - (p + 1)x + 1 = 0.Using mathematical induction show that \alpha^n \, + \, \beta^n\alpha^n \, + \, \beta^n is not divisible by p.

Let p\geq\geq 3 be an integer and \alpha\alpha , \beta\beta be the roots of x^2x^2 - (p + 1)x + 1 = 0.Using mathematical induction show that \alpha^n \, + \, \beta^n\alpha^n \, + \, \beta^n is an integer

Prove that $2^n>n$ for all positive integers $n$

For any natural number n , prove that inequality
|sin n x| $\geq$ n | sin x |

Prove that 7 is a factor of $2^{3n}$ - 1 for all natural numbers n.

Prove the assertions of the following problems
Prove that $2^n > n^2$ for any natural number $n > 5.$

If $P(n)$ is the statement $''2^n\ge 3n"$ and if $P(r)$ is true prove that $P(r+1)$ is true.

Using the principle of mathematical induction prove that $2 + 4 + 6 + .... + 2n = {n^2} + n$ .

Prove by mathematical induction,
${ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+....+{ n }^{ 2 }=\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 }$

Using the principle of mathematical induction, show that;
$2+4+6+.....+2n=n^2+n$

Using Principle of mathematical induction prove that $6^n -1$ divisible by $5$ .

Using the principle of Mathematical Induction, prove the following for all $n$ and $N$
1) ${3}^{n}>{2}^{n}$

If $P(n)$ is the statement $n^{2}-n+41$ is prime, prove that $P(1),P(2)$ and $P(3)$ are true.
Prove also that $P(41)$ is not true.

Solve: $(2n+7)<(n+3)^2$

Using P.M.I. prove that
${n^2} < {2^n}\forall n \ge 5$

Use the principle of mathematical induction to show that :
$2 + 2^2 + ..... + 2^n = 2^{n+1} - 2$ for every natural number $n.$

Show that only one of the no $n, n + 2$ and $n + 4$ in divisible $3$ .

Prove by method of induction, for all $n \in$
$1^2 + 2^2 + 3^2 + ... + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}$ .

To prove: $n \le 2^{n}$ for $n\ in\ N$
$P(n):n \le 2^{n}.........$ for $n\ in\ N$

Can the number $6^{n}$ , where n is a natural number, end with digit 5? Give reasons.

For every natural number $n(>1)$ , prove that $4^{n}+15n-1$ is divisible by $9$ .

Prove by method of induction; for all $n \in N$ $3 + 7 + 11 + \ldots . . . +$ to $n$ terms $= n ( 2 n + 1 )$

Prove that $3^{n+1} > 3(n+1)$ .

Prove that $1 + 2 + 3 + ..... + n =$ $\displaystyle \frac {n(n\, +\, 1)}{2}$ .

Prove that $3^{n + 1} > 3(n + 1)$ .

Prove $1+3+5+\cdots+(2n-1)=n^2$

Prove $3^{n+1}>$ $3(n+1)$

Prove the following using the principle of mathematical induction for all $n\in N$ :
$1+3+{3}^{2}+........+{3}^{n-1}=\cfrac{({3}^{n}-1)}{2}$

Prove that $1 + 3 + 5 + ..... + (2n - 1) = n$ $^2$ .

Prove the following by using principle of mathematical induction for all $n\in N: 1.3+3.5+5.7+.......+(2n-1)(2n+1)=\cfrac{n(4{n}^{2}+6n-1)}{3}$

Prove the following by using the principle of mathematical induction for all $n\in N: 1.3+2.{3}^{2}+3.{3}^{3}+.....+n.{3}{^n}=\cfrac{(2n-1){3}^{n+1}+3}{4}$

Prove the following by using the principle of mathematical induction for all $n\in N:1.2.3+2.3.4+......+n(n+1)(n+2)=\cfrac{n(n+1)(n+2)(n+3)}{4}$

Prove the following by using the principle of mathematical induction for all $n \in N: \cfrac{1}{2}+\cfrac{1}{4}+\cfrac{1}{8}+.....+\cfrac{1}{{2}^{n}}=1-\cfrac{1}{{2}^{n}}$

Prove the following by using the principle of mathematical induction for all $n\in$ :
$1+\cfrac{1}{(1+2)}+\cfrac{1}{(1+2+3)}+..........+\cfrac{1}{1+2+3+.....n)}=\cfrac{2n}{(n+1)}$

Prove the following by using the principle of mathematical induction for all $n\in N: 1.2+2.3+3.4+......+n(n+1)=\left[ \cfrac { n(n+1)(n+2) }{ 3 } \right]$

Prove the following by using the principle of mathematical induction for all $n\in N$
${ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+.......+{ n }^{ 3 }={ \left[ \cfrac { n(n+1) }{ 2 } \right] }^{ 2 }$

Prove the following by using the principle of mathematical induction for all $n\in N: 1.2+2.{2}^{2}+3.{2}^{2}+.......+n.{2}^{n}=(n-1){2}^{n+1}+2$

Prove the following by using the principle of mathematical induction for all $n\in N: \cfrac{1}{2.5}+\cfrac{1}{5.8}+\cfrac{1}{8.11}+......+\cfrac{1}{(3n-1)(3n+2)}=\cfrac{n}{(6n+4)}$

Prove the following by using the principle of mathematical induction for all $n\in N:{10}^{2n-1}+1$ is divisible by $11$

Prove the following by using the principle of mathematical induction for all $n\in N:{x}^{2n}-{y}^{2n}$ is divisible by $x+y$ .

Prove the following b y using the principle of mathematical induction for all $n\in N:1+2+3+.....+n<\cfrac{1}{8}{(2n+1)}^{2}$

Prove the following by using the principle of mathematical induction for all $n\in N: P(n):a+ar+a{r}^{2}+......+a{r}^{n-1}=\cfrac{a({r}^{n}-1)}{r-1}$

Prove the following by using the principle of mathematical induction for all $n\in N:{ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+.......+({ 2n-1) }^{ 2 }=\cfrac { n(2n-1)(2n+1) }{ 3 }$

Prove the following by using the principle of mathematical induction for all $n\in N:n(n+1)(n+5)$ is multiple of $3$ .

Find counter examples to disprove the following statement.
A quadrilateral with all sides are equal is a square.

Using mathematical induction, prove that $\cfrac{d}{dx}({x}^{n})=n{x}^{n-1}$ for all positive integers $n$ .

Prove the following by using the principle of mathematical induction for all $n\in N:{41}^{n}-{14}^{n}$ is a multiple of $27$ .

Prove the following by using the principle of mathematical induction for all $n\in N:(2n+7)< {(n+3)}^{2}$ .

Prove the following by using the principle of mathematical induction for all $n\in N:{3}^{2n+2}-8n-9$ is divisible by $8$ .

Prove by induction:
${1}^{2}+{2}^{2}+{3}^{2}+........+{n}^{2}=\cfrac{1}{6}n(n+1)(2n+1)$

Prove by induction:
${x}^{n}-{y}^{n}$ is divisible by $x+y$ when $n$ is even.

Show that ${ 2 }^{ 2n }-3n-1$ is divisible by $9$ for all positive integral values of $n$ .

Prove by induction:
$2+{2}^{2}+{2}^{3}+........+{2}^{n}=2({2}^{n}-1)$

Prove the following by the principle of mathematical induction: $(n\epsilon N)$
$1^{2} . 2 + 2^{2} . 3 + .... + n^{2} (n + 1) = \dfrac {n(n + 1)(n + 2)(3n + 1)}{12}$ .

Prove the following by the principle of Mathematical Induction.
$(2n + 1)(2n - 1)$ is an odd number for all $n\epsilon N$ .

Prove by induction:
$\cfrac{1}{1.2}+\cfrac{1}{2.3}+\cfrac{1}{3.4}+.....$ to $n$ terms $=\cfrac{n}{n+1}$

Prove by induction:
$1+3+5+.....+(2n-1)={n}^{2}$

If $p$ and $q$ are any two positive integers, show that $|\underline {pq}$ is divisible by $(|\underline {p})^{q} . |\underline {q}$ and by $(|\underline {q})^{p} . |\underline {p.}$

$(1 + x)^n >1$ + nx for n > 2 , n $\epsilon$ N
x > -1 x $\neq$ 0

Prove $2 + 5 + 8 + 11 + ... + (3n - 1) =$ $\dfrac{1}{2}$ $n (3n + 1) n$ $\epsilon$ N

Prove that $11^{n +2} + 12^{2n \, + \, 1}$ is divisible by $133$ for any non-negative integral $n$ .

Prove $\dfrac {1}{1 \cdot 2\cdot 3 } \, + \, \dfrac {1}{2 \cdot 3\cdot 4 } \, + \, \dfrac {1}{n (n + 1)(n + 2)} \, = \, \dfrac{1(n + 3)}{4(n + 1)(n + 2)}$

$\sqrt 2 + \sqrt 2 + \sqrt 2 + ...+ n\ terms = 2\cos \dfrac{\pi}{2^{n +1}} , n \epsilon N$

Prove the assertions of the following problems
Prove that the expression $n^3 \, - \, n$ is divisible by $24$ for any odd $n$ .

For any natural number n > 1
$\dfrac {1}{n \, + \, 1} \, + \, \dfrac {1}{n \, + \, 2} \, + \, ..... \, + \, \dfrac {1}{2n} > \dfrac {13}{14}$

Prove that induction that
sin x + sin 3x + .. + sin(2n - 1) x = $\dfrac {sin^2 nx}{sin x}$

Observing that $1^3 \, = \, 1, \, 2^3 \, = \, 3 \, + \, 5,$
$3^3$ = 7 + 9 + 11, $4^3$ = 13 + 15 + 17 + 19
Find a general formula for the cube of natural number n and prove it by the principle of mathematical induction

Explain the method of mathematical induction and use it to show that $11^{n \, + \, 2} \, + \, 12^{2n \, + \, 1}$ where n is natural number is divisible by 133

If $x_1x_2x_3 \, .... \, x_n \ = \, 1(x_1 \, > \, 0, i \, = \, 1,2,.....n),$ prove that $x_1 \, + \, x_2 \, + \, .....x_n \, \geq \, n \, (n \, \geq \, 2)$

Prove $^{2n} C _n \, > \, \dfrac{4^n }{n + 1}$

Prove $\dfrac {2n!}{(n!)^2} \, > \, \dfrac {4n}{2n \, + \, 1}$

If 0 < $\alpha < \, \dfrac{\pi}{4(n \, - \, 1)}$ where n > 1 , then prove that $\tan n$ $\alpha \, > \, n \, \tan \, \alpha$

Show by the mathematical induction that
$\dfrac{1}{sin 2 x } \, + \, \dfrac{1}{sin 4 x } \, + \, \dfrac{1}{sin 2^n x } \, = \, cot x \, - \, cot 2^n \, x$

Prove that ${ 1 }^{ 2 }+{ 2 }^{ 2 }+...+{ n }^{ 2 }>\cfrac { { n }^{ 3 } }{ 3 } ,n\epsilon N$

Prove the following by using the first principle
$1+3 + 3^2 + ......+ 3^{n-1} =\frac{(3^n -1)}{2}$ .

Prove the following by PMI for all $n$ belong to $N$
$1\times3+3\times5+5\times7 .........+(2n-1)(2n+1)=\dfrac{n(4n^2+6n-1)}{3}$

Prove that for any positive interger $n, n^3 -n$ is divisible by $6$ .

Let p $\geq$ 3 be an integer and $\alpha$ , $\beta$ be the roots of $x^2$ - (p + 1)x + 1 = 0.Using mathematical induction show that $\alpha^n \, + \, \beta^n$
is not divisible by p.

Let p $\geq$ 3 be an integer and $\alpha$ , $\beta$ be the roots of $x^2$ - (p + 1)x + 1 = 0.Using mathematical induction show that $\alpha^n \, + \, \beta^n$
is an integer

Let $I_m \, = \, \int_{0}^{\pi}\dfrac{1 \, - \, cos \, \, mx}{1 \, - \, cos \, \, x}dx$
Use mathematical induction to prove that
$I_m \, = \, m\pi \, , \, m0$ , , 1, 2.....

By PMC, prove that inequality $n<2^n$ for all $n\in N$

Use mathematical induction to prove
$1 + 2 + 3 + \ldots + n < \frac{1}{8}{\left( {2n + 1} \right)^2}$