MCQ Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 1

Statement-l: For every natural number

$n\geq 2,\ \displaystyle \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots\ldots+\frac{1}{\sqrt{n}}>\sqrt{n}$ .
Statement-2: For every natural number

$n\geq 2,\ \sqrt{n(n+1)}<n+1$ .

0%
Statement-1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1.
0%
Statement-1 is true, Statement-2 is false.
0%
Statement-1 is false, Statement-2 is true.
0%
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Explanation

$\displaystyle P\left( n \right) =\frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 2 } } +...+\frac { 1 }{ \sqrt { n } }$

$\displaystyle P\left( 2 \right) =\frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 2 } } >\sqrt { 2 }$
Let us assume that

$P(k)$

$\displaystyle =\frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 2 } } +...+\frac { 1 }{ \sqrt { k } } >\sqrt { k }$ is true

$\displaystyle \therefore P\left( k+1 \right) =\frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 2 } } +...+\frac { 1 }{ \sqrt { k } } +\frac { 1 }{ \sqrt { k+1 } } >\sqrt { k+1 }$ has to be true.

$\displaystyle L.H.S.>\sqrt { k } +\frac { 1 }{ \sqrt { k+1 } } =\frac { \sqrt { k\left( k+1 \right)} +1 }{ \sqrt { k+1 } }$
Since

$\sqrt { k\left( k+1 \right) } >k\quad \left( \forall k\ge 0 \right)$

$\displaystyle \therefore \frac { \sqrt { k\left( k+1 \right) }+1 }{ \sqrt { k+1 } } >\frac { k+1 }{ \sqrt { k+1 } } =\sqrt { k+1 }$
Let

$P\left( n \right) =\sqrt { n\left( n+1 \right) } <\left( k+1 \right)$
Statement-1 is correct.

$P\left( 2 \right) =\sqrt { 2\times 3 } <3$
If

$P\left( k \right) =\sqrt { k\left( k+1 \right) } <\left( k+1 \right)$ is true
Now

$P\left( k+1 \right) =\sqrt { \left( k+1 \right) \left( k+2 \right)} <k+2$ has to be true
Since

$\left( k+1 \right) <k+2$

$\therefore \sqrt { \left( k+1 \right) \left( k+2 \right) } <\left( k+2 \right)$
Hence Statement-2 is not correct explanation of Statement-1.

Let

$S(k) = 1 + 3 + 5 + .... + (2k - 1) = 3 + k^2$ . Then which of the following is true?

0%
Principle of mathematical induction can be used to prove the formula
0%
$S (k)$ $\Rightarrow$ $S (k + 1)$
0%
$S (k)$ ${\nRightarrow}$ $S (k + 1)$
0%
$S (1)$ is correct

Explanation

Putting

$k=1$ , we get L.H.S=1 and R.H.S=4. Hence

$A$ and

$D$ are incorrect.

Now,

$S\left( k+1 \right) =1+3+5+$ ...

$+\left( 2k-1 \right) +\left( 2\left( k+1 \right) -1 \right)$

$\Rightarrow S\left( k+1 \right) =S\left( k \right) +\left( 2k+1 \right)$ [Since

$S\left( k \right) =1+3+5+$ ...

$+\left( 2k-1 \right)$ ]

$\Rightarrow S\left( k+1 \right) =3+{ k }^{ 2 }+\left( 2k+1 \right)$

$\Rightarrow S\left( k+1 \right) =3+{ \left( k+1 \right) }^{ 2 }$

$B$ is correct option.

Mathematical Induction is the principle containing the set

0%
R
0%
N
0%
Q
0%
Z

Let

$P(n)$ be a statement and

$P(n)=P(n+1) \forall n\in N$ , then

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

$n$ ?

0%
For all $n$
0%
For all $n>1$
0%
For all $n>m$ , $m$ being a fixed positive integer
0%
Nothing can be said

Explanation

Given,

$P(n) = P(n+1) \forall n\in N$
Substituting

$n-1$ in place of

$n$ ,

$P(n-1)=P(n)$
Thus if

$P(k)$ is true for some

$k$

$\in$

$N$ , then it is true for

$k-1$ and

$k+1$ .
Thus, it is true

$\forall k$

$\in$

$N$

State whether the following statement is true or false.
cos x + cos 2x + .... + cos nx =

$\dfrac{cos\left (\dfrac{n \, + \, 1 }{2} \right )x sin \dfrac{nx}{2}}{sin\dfrac{x}{2}}$

0%
True
0%
False

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

0%
True
0%
False

Explanation

Let

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

Step 1 :

Put

$n=1$

Then, LHS = 1

RHS =

$1^2=1$

Therefore, LHS = RHS

$\implies P(n)$ is true for

$n=1$ .

Step 2 :

Assume that

$P(n)$ is true 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,

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

$\implies P(n)$ is true for

$n=k+1$ .

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

A bag contains

$3$ red and

$2$ black balls. One ball is drawn from it at random. Find the probability of drawing red ball is

$\dfrac 35$

0%
True
0%
False

Explanation

No o.f red balls

$=3$

NO.off Black balls

$=2$

Total no .of balls

$=2+3=5$

Probability

$=\dfrac{3}{5}$

Let

$P(n)$ be the statement

$"3^n>n"$ . If

$P(n)$ is true,

$P(n+1)$ is true.

0%
True
0%
False

Explanation

$P(n)$ is true.

$3^n>n$

$3.3^n>3n$

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

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

$[ 2n>1$ for every

$n\in N \implies 2n+n >n+1$ for every

$n\in N]$

$P(n+1)$ is true.

Let

$P(n)= 5^{n}-2^{n}$ .

$P(n)$ is divisible by

$3\lambda$ where

$\lambda$ and

${n}$ both are odd positive integers, then the least value of

$n$ and

$\lambda$ will be

0%
$13$
0%
$11$
0%
$1$
0%
$5$

Explanation

$5^n-2^n$ is divisible by

$5-2=3$ always... Putting

$n=\lambda =1$ which is the least odd positive integer, this works to be true.
Hence Option C

For every integer

$n\geq 1, (3^{2^{n}}-1)$ is always divisible by

0%
$2^{n^2}$
0%
$2^{n+4}$
0%
$2^{n+2}$
0%
$2^{n+3}$

Explanation

For

$n= 1$ ,

$3^{2^{1}}-1 = 8$ , which is divisible by

$2^{n+2}$ .

Let us assume that

$3^{2^m} -1$ is divisible by

$2^{m+2}$ for some integral value of

$m$ .
Let us consider the expression for

$m+1$

$3^{2^{m+1}} -1$

$= (3^{2^{m}} -1) \times (3^{2^{m}} +1)$
The first term is divisible

$2^{m+2}$ and the second term is also an even number.
Hence, the term is divisible by

$2^{m+2}$ .
Hence, by induction we can prove that it is true for all

$m$ .

$\forall n\in N; x^{2n-1}+y^{2n-1}$ is divisible by?

0%
$x-y$
0%
$x+y$
0%
$xy$
0%
$x^{2}+y^{2}$

Explanation

$\displaystyle P\left ( n \right )=x^{2n-1}+y^{2n-1}\forall n \epsilon N.$
Substitute

$n=1$ to obtain

$\displaystyle p\left ( 1 \right )= x+y$ ,Which is divisible by

$x+y$ .
For,

$n=2$ , we get

$\displaystyle P\left ( 2 \right )=x^{3}+y^{3}=\left ( x+y \right )\left ( x^{2}-xy+y^{2} \right )$ which is divisible by

$x+y$ .
With the help of induction we conclude that

$P(n)$ will be divisible by

$x+y$ for all

$n\in N$ .

Ans: B

Let

$\mathrm{S}(\mathrm{K})=1+3+5+\ldots\ldots..+(2\mathrm{K}-1)=3+\mathrm{K}^{2}$ . Then which of the following is true?

0%
$\mathrm{S}(1)$ is correct
0%
$\mathrm{S}(\mathrm{K})\Rightarrow \mathrm{S}(\mathrm{K}+1)$
0%
$S(\mathrm{K})\neq$ S(K $+$ 1)
0%
Principle of mathematical induction can be used to prove the formula

Explanation

$S\left( K \right) =1+3+5+...+\left( 2K-1 \right) =3+{ K }^{ 2 }$

Put

$K=1$ in both sides

$\therefore L.H.S=1$ and

$R.H.S=3+1=4$

$\Rightarrow L.H.S\neq R.H.S$

Put

$\left( K+1 \right)$ on both sides in the place of

$k$

$L.H.S=1+3+5+...+\left( 2K-1 \right) +\left( 2K+1 \right)$

and

$R.H.S=3+{ \left( K+1 \right) }^{ 2 }=3+{ K }^{ 2 }+2K+1$

Let

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

$\Rightarrow 1+3+5+...+\left( 2K-1 \right) +\left( 2K+1 \right) =3+{ K }^{ 2 }+2K+1\\ \Rightarrow 1+3+5+...+\left( 2K-1 \right) =3+{ K }^{ 2 }$

$S(K)$ is true, then

$S(K+1)$ is also true.

Hence

$S\left( K \right) \Rightarrow S\left( K+1 \right)$

$n(n^{2}-1)$ is divisible by

$24$ , then which of the following statements is true?

0%
$n$ can be any odd integral value.
0%
$n$ can be any integral value.
0%
$n$ can be any even integral value.
0%
$n$ can be any rational number.

Explanation

$n({ n }^{ 2 }-1)\quad =\quad n*(n-1)*(n+1)$
If

$n$ is even, then

$n-1$ and

$n+1$ will be odd, therefore

$n({ n }^{ 2 }-1)$ is not divisible by 4 and therefore not divisible by 24.
Hence

$n$ has to be an odd integer.

$\forall m\in N$ , then

$11^{m+ 2}+12^{2m-1}$ is divisible by

0%
$121$
0%
$132$
0%
$133$
0%
None of these

Explanation

To find the divisor of

${ 11 }^{ m+2 }+{ 12 }^{ 2m-1 }$ by mathematic induction, the first step is to check for the smallest natural number, i.e; for

$m=1$ . So, this reduces to

${ 11 }^{ 3 }+{ 12 }^{ 1 }$ or

${ 11 }^{ 4 }+1$ .

So, the number when divided by

$11$ leaves remainder 1.

So, we can knock out options

$A$ and

$B$ as

$121$ as well as

$132$ are both divisible by 11 and hence their multiples will always be divisible by 11.

Now, we have to check the divisibility of

${ 11 }^{ m+2 }+{ 12 }^{ 2m-1 }$ by

$133$ . For

$m=1$ ,

${ 11 }^{ 4 }+1$ is not divisible by

$133$ .

So, we can knock out option

$C$ .

Hence,

$D$ is correct.

$n$ is an even number, then the digit in the units place of

$2^{2n}+1$ will be

0%
$5$
0%
$7$
0%
$6$
0%
$1$

Explanation

Since

${ 2 }^{ 2n }$ is even therefore

${ 2 }^{ 2n }+1$ is odd,
therefore digit at unit place should be odd, rejecting option 3.
Put n = 2, we get

${ 2 }^{ 2n } +1$ = 17,

hence digit should be 7

If A =

$\begin{vmatrix} 1 &0 \\ 1& 1 \end{vmatrix}$ B and I =

$\begin{vmatrix} 1 &0 \\ 0& 1 \end{vmatrix}$ ,then which one of the following holds for all n

$\geq$ 1, by
the principle of mathematical indunction

0%
$A^{n}=nA-(n-1)l$
0%
$A^{n}=2^{n-1}A-(n-1)l$
0%
$A^{n}=nA+(n-1)l$
0%
$A^{n}=2^{n-1}A+(n-1)l$

Explanation

$A=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\\ { A }^{ 2 }=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\\ { A }^{ 3 }=\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}\\ \therefore { A }^{ n }=\begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix}$

and

$nA=n\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} n & 0 \\ n & n \end{bmatrix}$

$\left( n-1 \right) I=\left( n-1 \right) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} n-1 & 0 \\ 0 & n-1 \end{bmatrix}\\ \therefore nA-\left( n-1 \right) I=\begin{bmatrix} n & 0 \\ n & n \end{bmatrix}-\begin{bmatrix} n-1 & 0 \\ 0 & n-1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix}={ A }^{ n }$

Let

$P(n):1+\displaystyle \frac{1}{4}+\frac{1}{9}+\ldots..+\frac{1}{n^{2}}<2-\frac{1}{n}$ is true for

0%
$\forall n\in N$
0%
$n=1$
0%
${n>1,\forall n\in N}$
0%
$n>2$

Explanation

For

$n=2,P\left( n \right) =1+\dfrac { 1 }{ { 2 }^{ 2 } } =1+\dfrac { 1 }{ 4 } =\dfrac { 5 }{ 4 } =1.25<2-\dfrac { 1 }{ 2 } =\dfrac { 3 }{ 2 } =1.5\\ P(3)=1+\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 9 } =1.36<1.67$

$\therefore P(2), P(3)$ is true and so on.

Let us assume

$P(n)$ is true.

$P\left( n+1 \right) =1+\dfrac { 1 }{ 4 } +.....+\dfrac { 1 }{ { n }^{ 2 } } +\dfrac { 1 }{ { \left( n+1 \right) }^{ 2 } } <2-\dfrac { 1 }{ n } +\dfrac { 1 }{ { \left( n+1 \right) }^{ 2 } } \\ =2-\left\{ \dfrac { 1 }{ n } -\dfrac { 1 }{ { \left( n+1 \right) }^{ 2 } } \right\} =2-\dfrac { { n }^{ 2 }+n+1 }{ n{ \left( n+1 \right) }^{ 2 } } =2-\dfrac { { n }^{ 2 }+n }{ n{ \left( n+1 \right) }^{ 2 } } -\dfrac { 1 }{ n{ \left( n+1 \right) }^{ 2 } }$

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

$\therefore P(n+1)$ is true.

Thus,

$P(n)$ is true for all

$n>1.$

Let

$x > -1$ , then statement

$p(n):(1 + x)^{n} > 1 + nx$ , where

$n \in N$ is true for

0%
For all $n \epsilon N$ .
0%
For all $n > 1$ .
0%
For all $n > 1$ , provided $x \neq 0$ .
0%
For all $n > 2$ .

Explanation

p(1) :

$(1 + x)^{1}$ > 1 + x is false
p(2) :

$(1 + x)^{2} > 1 + 2x \Rightarrow x^{2}$ > 0 is true when

$x \neq 0$
p(3) :

$(1 + x)^{3} > 1 + 3x \Rightarrow x^{2}$ > 0 is true when

$x \neq 0$
Let p(k)

$(1 + x)^{k}$ > 1 + k x is true for some

$k \epsilon N$ , k >1

$\Rightarrow (1+x)^{k+1}$ >(1+kx) (1+x)

$\Rightarrow (1+x)^{k+1} > 1+(k+1)x+kx^{2} > 1 +(k+1)x, x \neq 0$

$\Rightarrow$ p(k+1) is true whenever p(k) is true
so by principle of mathematical induction p(n) is true for all n > 1 provided

$x \neq 0$

$1.2 +2.2^2 +3.2^3+ .................+ n.2^n = (n-1)2^{n+1} + 2$ is true for

0%
Only natural number $n$ $\geq$ $3$
0%
All natural number $n$
0%
Only natural number $n$ $\geq$ $5$
0%
None

Explanation

Let P(n) be the given statement i.e.

$P(n) : 1.2 + 2.2^2 + 3.2^3+ ......... + n.2^n = (n-1)2^{n+1}+2$
Putting n

$=$ 1, LHS

$=$ 1.2

$=$ 2; RHS

$=$ 0+2

$=$ 2

$\therefore$ P(n) is true for

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

$n=k$
i.e. P(k) is true i.e.

$1.2^2+ 2.2^2 + 3.2^3 + ........... + k.2$

$^k$ Replacing k by k + 1, we get the next term

$=(k+1)2^{k+1}$
Adding it to both sides

$LHS = 1.2 +2.2^2 + 3.2^3+ ......... + k.2^{k}+ (k+1)2^{k+1}$

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

$=2^{k+1}[k-1+k+1]+2=2k2^{k+1}+2=(k+1-1)2^{k+1+1}+2$

This proves P(n) true for

$n= k+1$
Thus P(k+1) is true whenever P(k) is true
Hence. P(n) is true for all n

$\in$ N

$n(n+1) (n+5)$ is a multiple of

$3$ is true for

0%
All natural numbers $n > 5$
0%
Only natural number $3$ $\leq$ $n < 15$
0%
All natural numbers $n$
0%
None

Explanation

Let the statement be denoted by p(n) i.e.,
P(n) : n(n+1) (n+5) is a multiple of 3
For n

$=$ 1, n(n+1) (n+5)

$=$ 1.2.6

$=$ 12

$=$ 3.4
P(n) is true for n

$=$ 1
Suppose p(k) is true for n

$=$ k i.e.
k(k+1) (k+5) =3m (let) or k

$^3$ + 6k

$^2$ + 5k

$=$ 3m ........... (i)
Replacing k by k+1, we get
(k+1) (k+2) (k+6)

$=$ k (k

$^2$ +8k +12) + (k

$^2$ + 8k + 12)

$k^3 + 9k^2 +20 k +12 = (k^3 + 6k^2 + 5k) + (3k^2 +15k + 12)$

$= 3m + 3k^2 +15k + 12$ [from (i)]

$= 3(m + k^2 + 5k +4)$
i.e. (k+1) (k+2) (k+6) is a multiple of 3
i.e. P(k+1) is multiple of 3, if P(k) is a multiple of 3
i.e. P(k+1) is true whenever P(k) is true.
Hence P(n) is true for all n

$\in$ N

$n\in N$ , then

$n(n^2-1)$ is divisible by

0%
$6$
0%
$16$
0%
$26$
0%
$24$

Explanation

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

One of the

$n$ ,

$n+1$ and

$n-1$ will be a multiple of

$3$ .

Since

$n-1$ ,

$n$ and

$n+1$ are three consecutive integers, therefore at least one of them will be divisible by

$2$ .

Therefore

${ n }{ (n }^{ 2 }-1)$ is divisible by

$6$ .

Option d can be rejected by putting

$n = 2$

Using mathematical induction,

$\displaystyle \left ( 1 - \frac{1}{2^2} \right ) \left ( 1 - \frac{1}{3^2} \right ) \left ( 1 - \frac{1}{4^2} \right ) ......... \left ( 1 - \frac{1}{(n + 1)^2} \right )$

0%
$\displaystyle \frac{n +2}{2(n + 1)}$
0%
$\displaystyle \frac{n - 2}{2 (n - 1)}$
0%
$\displaystyle \frac{n + 3}{2 (n + 1)}$
0%
$\displaystyle \frac{n}{2 (n + 1)}$

Explanation

$\displaystyle \left ( 1 - \dfrac{1}{2^2} \right ) \left ( 1 - \dfrac{1}{3^2} \right ) \left ( 1 - \dfrac{1}{4^2} \right ) ......... \left ( 1 - \dfrac{1}{(n + 1)^2} \right )$

nth term is

$1-\dfrac{1}{(n+1)^2}$

for n=1 =>

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

$1-\dfrac{1}{4}$ =

$\dfrac{3}{4}$

for n-1 series will be

$\left( 1-\dfrac { 1 }{ 2^{ 2 } } \right) \left( 1-\dfrac { 1 }{ 3^{ 2 } } \right) \left( 1-\dfrac { 1 }{ 4^{ 2 } } \right) .........\left( 1-\dfrac { 1 }{ (n+1-1)^{ 2 } } \right) \left( 1-\dfrac { 1 }{ (n+1)^{ 2 } } \right)$

On simplyfying

$\left( \dfrac { 3 }{ { 2 }^{ 2 } } \right) \left( \dfrac { 8 }{ { 3 }^{ 2 } } \right) \left( \dfrac { 15 }{ { 4 }^{ 2 } } \right) ........\left( \dfrac { (n+1)(n-1) }{ { n }^{ 2 } } \right) \left( \dfrac { n(n+2) }{ { (n+1) }^{ 2 } } \right)$

$\left( \dfrac { 3 }{ { 2 }^{ 2 } } \right) \left( \dfrac { 2*4 }{ { 3 }^{ 2 } } \right) \left( \dfrac { 3*5 }{ { 4 }^{ 2 } } \right) ........\left( \dfrac { (n+1)(n-1) }{ { n }^{ 2 } } \right) \left( \dfrac { n(n+2) }{ { (n+1) }^{ 2 } } \right)$

Now all terms will cancel out with each other from denominator to numinator only

$\dfrac{1}{2}$ from first term and

$(\dfrac{n+2}{n+1})$ from last term will remain

Therefore, Answer is

$\dfrac{1}{2}(\dfrac{n+2}{n+1})$

The product of five consecutive natural numbers is divisible by

0%
10
0%
20
0%
30
0%
120

Explanation

Let

$k$ consecutive natural number be

$(n+1), (n+2)........,(n+k)$

Thus their product is,

$P = (n+1)(n+2)(n+3).........(n+k)$

$\Rightarrow P=\dfrac{n!(n+1)(n+2)........(n+k)}{n!} =\dfrac{(n+k)!}{n!k!}\times k!= ^{n+k}C_k\times k! =k!\times$ Integer

Hence the product of

$k$ natural number is always divisible by

$k!$

Here, given

$k=5$ , so

$k! =120$ which is divisible by

$10,20,30,120$

For all positive integers

$n$ ,

$P(n)$ is true , and

$2^{n-2}>3n$ , then which of the following is true?

0%
$P(3)$ is true.
0%
$P(5)$ is true.
0%
If $P(m)$ is true then $P(m + 1)$ is also true.
0%
If $P(m)$ is true then $P(m + 1)$ is not true.

Explanation

$2^{n-2}>3n$
Put

$n=3$

$2\ngtr 6$
Hence, P(3) is not true.

Put

$n=5$

$8\ngtr 15$
Hence, P(3) is not true.

Let

$P(m)$ is true i.e.

$2^{m-2}>3m$

Now, we will check for

$P(m+1)$
Consider ,

$2^{m-1}$

$=2^{m-2}.2$

$>2(3m)=6m=3m+3m$

$>3m+3$
Hence,

$2^{m-1}>3(m+1)$
Hence,

$P(m+1)$ is true.

$\displaystyle \frac{1}{2} + \frac{1}{4}+ \frac{1}{8} + ......... + \frac{1}{2^n} = 1 - \frac{1}{2^n}$ is true for

0%
Only natural number n $\geq$ 3
0%
Only natural number n $\geq$ 5
0%
Only natural number n < 10
0%
All natural number n

Explanation

Let P(n)

$: \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + .......... + \frac{1}{2^n} = 1 - \frac{1}{2^n}$
Putting

$n=1, LHS = \frac{1}{2} , RHS = 1 - \frac{1}{2} = \frac{1}{2}$ i.e., LHS

$=$ RHS

$= \frac{1}{2}$

$\therefore$ P(n) is true for n

$=$ 1.
Suppose P(n) is true for n

$=$ k

$\therefore \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ......... + \frac{1}{2^k} = 1 - \frac{1}{2^k}$
last term

$= \displaystyle \frac{1}{2^k}$ ; Replacing k by k+1, last term

$=\displaystyle \frac{1}{2^{k+1}}$
Adding

$\displaystyle \frac{1}{2^{k+1}}$ to both sides,

$LHS = \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ........ + \frac{1}{2^k}+ \frac{1}{2^{k+1}}$

$RHS = 1 - \displaystyle \frac{1}{2^k} + \frac{1}{2^{k+1}}=1-\dfrac{1}{2^k} \left ( 1 - \frac{1}{2} \right ) = 1 - \frac{1}{2^k} \cdot \frac{1}{2} = 1 - \frac{1}{2^{k+1}}$
This shows P(n) is true for n

$=$ k+1
Thus P(k+1) is true whenever P(k) is true
Hence, P(n) is true for all n

$\in$ N

$P(n)$ is statement such that

$P(3)$ is true. Assuming P(k) is true

$\Rightarrow$

$P(k+1)$ is true for all

$k$

$\geq$

$2$ , then

$P(n)$ is true.

0%
For all $n$
0%
For $n$ $\geq$ $3$
0%
For $n$ $\geq$ $4$
0%
None of these

Explanation

Given

$P (3)$ is true.
Assume

$P(k)$ is true

$\Rightarrow$

$P(k+1)$ is true means if

$P(3)$ is true

$\Rightarrow$

$P(4)$ is true

$\Rightarrow$

$P(5)$ is true and so on. So statement is true for all

$n\geq3$ .

$1.3 + 3.5 + 5.7 + ........... + (2n -1) (2n + 1) = \displaystyle \frac{n (4n^2 + 6n -1)}{3}$ is true for

0%
Only natural number n $\geq$ 4
0%
Only natural numbers 3 $\leq$ n $\leq$ 10
0%
All natural numbers n
0%
None

Explanation

Le P(n) be the given statement
i.e. P(n) : 1.3 + 3.5 + 5.7 + ........... + (2n -1)(2n+1)

$=\displaystyle \frac{n (4n^2 + 6n -1)}{3}$
Putting

$n=1, L.H.S. = 1.3 = 3$ and

$RHS = \displaystyle \frac{1 (4.1^2 + 6.1 -1)}{3}$

$= \displaystyle \frac{4 + 6 -1}{3} = \frac{9}{3} = 3$
LHS

$=$ RHS

$\therefore$ P(n) is true for n

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

$=$ k
i.e., p(k) is true
i.e., P(k) : 1.3 + 3.5 + 5.7 + ............ + (2k-1)(2k+1)

$= \displaystyle \frac{k (4k^2 + 6k -1)}{3}$
Last term

$=$ (2k -1)(2k +1)
Replacing k by (k+1), we get

$[2(k+1)-1] [2 (k+1)+1]= (2k+1)(2k + 3)$

$\therefore$ Adding (2k+1)(2k+3) on both sides.

$\therefore LHS = 1.3 + 3.5 + 5.7 + ........... + (2k-1) (2k +1) + (2k +1) (2k +3)$

$R.H.S = \displaystyle \frac{k (4k^2 + 6k -1)}3{} + (2k +1) (2k +3)$

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

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

$= \displaystyle \frac{(k+1) (4k^2 + 14 k +9)}{3}$

$= \displaystyle \frac{(k+1)(4 (k+1)^2 + 6 (k+1)-1)}{3}$ ............ (iii)
Thus, P(n) is true for n

$=$ k + 1

$\therefore$ P(k+1) is true whenever P(k) is true
Hence, by P(n) is true for all n

$\in$ N

$\forall n\in N, 2 \cdot 4^{2n + 1} + 3^{3n + 1}$ is divisible by

0%
$7$
0%
$5$
0%
$11$
0%
$209$

Explanation

For

$n = 1$ , we have

$2 \cdot 4^{2n + 1} + 3^{3n + 1} = 2 \times 4^3 + 3^4 = 209$ , which is divisible by

$11$
For

$n = 2$ , we have

$2 \cdot 4^{2n + 1} + 3^{3n + 1} = 2 \times 4^5 + 3^7 = 4235$ , which is divisible by

$11$
Hence, options (c) is true.

$\forall n\in N, 3^{3n} - 26^{n}$ is divisible by

0%
$24$
0%
$64$
0%
$17$
0%
none of these

Explanation

Put

$n=2$

$3^{3n}-26^n = 3^6-26^2 =53$ which is not divisible by any

$24,64,17$
Hence option 'D' is correct choice.

$7^{2n} + 3^{n - 1} \cdot 2^{3n - 3}$ is divisible by

0%
$24$
0%
$25$
0%
$9$
0%
$13$

Explanation

Let

$P(1) = 7^{2n} + 3^{n - 1} \cdot 2^{3n - 3}$

$P(1) = 50\Rightarrow$ Divisible by

$25$

Hence option 'B' is the correct choice.

CBSE Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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