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

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

Putting $$k=1$$, we get L.H.S=1 and R.H.S=4. Hence $$A$$ and $$D$$ are incorrect.

Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers.

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 $$\forall k$$ $$\in$$ $$N$$ 

Using mathematical induction we can prove it.

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

$$P(n)$$ is true.

$$5^n-2^n$$ 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 $$n= 1$$, $$3^{2^{1}}-1 = 8 $$ , 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 $$m$$.

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

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

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

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  $${ 11 }^{ 4 }+1$$.

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, 

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

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 $$x \neq  0$$

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

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

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

Let $$k$$ consecutive natural number be  $$(n+1), (n+2)........,(n+k)$$

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

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

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 $$n\geq3$$.

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

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.

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.

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