MCQ Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 4

A student was asked to prove a statement by induction.
(i)

$P(5)$ is true and
(ii) Truth of

$P(n) \Rightarrow$ truth of

$P(n + 1)$ ,

$n\epsilon N$
On the basis of this, he could conclude that

$P(n)$ is true for

0%
no $n\epsilon N$
0%
all $n\epsilon N$
0%
all $n\geq 5$
0%
none of these

$n \in N$ , then

$x^{2n+1} + y^{2n+1}$ is divisible by

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

The inequality

$n! > 2^{n-1}$ is true

0%
for all $n > 1$
0%
for all $n > 2$
0%
for all $n \epsilon N$
0%
none of these

Explanation

Given inequality,

$n!>2^{n-1}$

For

$n=1$ ,

$LHS=n!=1!=1$

$RHS=2^{n-1}=2^{1-1}=2^0=1$

Inequality not satisfied.

For

$n=2$ ,

$LHS=n!=2!=2\times1=2$

$RHS=2^{n-1}=2^{2-1}=2^1=2$

Inequality not satisfied.

For

$n=3$ ,

$LHS=n!=3!=3\times2\times1=6$

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

Inequality is satisfied in this case.

Hence, the inequation is true for

$n>2$ .

For positive integer n,

$10^{n-2} > 81n$ when

0%
$n < 5$
0%
$n > 5$
0%
$n\geq 5$
0%
$n > 6$

Explanation

$10^{n-2}>81n$

For

$n=5$ ,

LHS=

$10^{5-2}=10^3=1000$

RHS=

$81n=81\times5=405$

$\Rightarrow10^{n-2}>81n$ for all

$n\ge5$ . Hence, the answer.

State which of the following statements is true

$\displaystyle 2^{16}-1$ is divisible by

0%
11
0%
13
0%
17
0%
19

Explanation

$\displaystyle ^{16}-1=\left ( 2^{4} \right )^{4}-1=\left ( 16 \right )^{4}-1$
=(16-1)(16+1)(16

$\displaystyle ^{2}$ +1)
Hence 2

$\displaystyle ^{16}$ -1 is divisible by 17

$7$ is a factor of

$2^{3n}-1$ for all natural numbers n.

0%
True
0%
False

Explanation

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) is true for

$n=1$
Step 2: Let P(n) be true for

$n=k$

$\implies$ 7 is a factor of

$2^{3k}-1$ .

$\implies2^{3k}-1=7M$ , where

$M\in N$ .

$\implies2^{3k}=7M+1\rightarrow(1)$
Now consider

$2^{\displaystyle3(k+1)}-1=2^{\displaystyle3k+3}-1=2^{3k}.2^3-1$

$=8(7M+1)-1$ using (1)

$=56M+7$ (As

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

$\therefore2^{\displaystyle3(k+1)}-1=7(8M+1)$

$\implies$ 7 is a factor of

$2^{\displaystyle3(k+1)}-1$

$\implies$ 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, 7 is a factor of

$2^{3n}-1$

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%
$S(K) \not{\Rightarrow} S(K + 1)$
0%
S(1) is correct

Explanation

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

$S(1) = 1= 3+ 1,$ which is not true

$\because$ S(1) is not true.

$\therefore$ P.M. 1 cannot be applied

Let

$S(k)$ is true, i.e.

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

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

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

$\therefore S(k) \Rightarrow S(k + 1)$

The statement P(n)

$"1 \times 1! + 2\times 2! + 3\times 3! + ... + n \times n! =(n + 1)! 1"$ is

0%
True for all n > 1
0%
Not true for any n
0%
True for all $n\epsilon N$
0%
None of these

Explanation

$P\left( 1 \right) :1\times 1=\left( 1+1 \right) =\left( 1+1 \right) !-1$ is true

Let

$p(m)$ be true

$\Rightarrow 1\times 1!+2\times 2!+3\times 3!+...+m\times m!=\left( m+1 \right) !-1$

$\Rightarrow 1\times 1+2\times 2!+3\times 3!+...+m\times m!+\left( m+1 \right) \times \left( m+1 \right) !$

$=\left( m+1 \right) !-1\left( m+2 \right) \left( m+1 \right) !=\left( m+1 \right) !\left( m+2 \right) -1$

$=\left( m+2 \right) !-1$

$\Rightarrow p(m+1)$ is also true

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

$n$

$a, b$ and

$n$ are natural numbers then

$a^{2n-1}+b^{2n-1}$ is divisible by

0%
$a + b$
0%
$a - b$
0%
$a^3+b^3$
0%
$a^2+b^2$

Explanation

Let

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

$P(1)=a+b$

$P(2)=a^3+b^3=(a+b)(a^2+b^2-ab)$

Out of given options we can conclude that

$P(n)$ is divisible only by

$(a+b)$

It can also be done by Mathematical Induction.

But here it is objective, so verification is good enough.

Let

$a, b, c$ and

$d$ be any four real numbers. Then,

$a^n+b^n=c^n+d^n$ holds for any natural number

$n$ , if

(This question has some ambiguity, but appeared in WBJEE 2015 exam).

0%
$a+b = c+d$
0%
$a-b=c-d$
0%
$a+b=c+d, a^2+b^2=c^2+d^2$
0%
$a-b=c-d, a^2-b^2=c^2-d^2$

Explanation

Option A

$a+b=c+d$ ....Given

Squaring on both sides, we get

$(a+b)^2 = (c+d)^2$

$a^2 +b^2 + 2ab = c^2+d^2+2cd$

$\therefore a^n+b^n \neq c^n+d^n$

Option B

$a-b=c-d$ ...Given

Squaring on both sides, we get

$(a-b)^2 = (c-d)^2$

$a^2 +b^2 - 2ab = c^2+d^2-2cd$

$\therefore a^n+b^n \neq c^n+d^n$

Option C

$a+b=c+d$ ...(i)

$a^2+b^2 = c^2+d^2$ ....(ii)

Consider

$(a+b)^2 = (c+d)^2$

$a^2 +b^2 + 2ab = c^2+d^2+2cd$

$\Rightarrow 2ab = 2cd$ ...from (ii)

$\Rightarrow ab=cd$

So,

$a^n+b^n \neq c^n+d^n$ for all

$n \epsilon N$ .

Option D

$a-b=c-d$ ...(i)

$a^2-b^2 = c^2-d^2$ ....(ii)

Consider

$a^2-b^2 = c^2-d^2$

$\Rightarrow (a-b)(a+b) = (c-d)(c+d)$

$\Rightarrow a+b = c+d$ ...from (i) ....(iii)

Adding eq(i) and (ii), we get

$2a=2c$

$\Rightarrow a=c$

$\therefore b=d$ ...from (iii)

So,

$a^n+b^n=c^n+d^n$ for all

$n \epsilon N$ .

For any integer

$n\ge 1$ , the sum

$\displaystyle\sum _{ k=1 }^{ n }{ k\left( k+2 \right) }$ is equal to

0%
$\dfrac { n\left( n+1 \right) \left( n+2 \right) }{ 6 }$
0%
$\dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 }$
0%
$\dfrac { n\left( n+1 \right) \left( 2n+7 \right) }{ 6 }$
0%
$\dfrac { n\left( n+1 \right) \left( 2n+9 \right) }{ 6 }$

Explanation

Now,

$\displaystyle\sum _{ k=1 }^{ n }{ k\left( k+2 \right) }$

$=\displaystyle\sum _{ k=1 }^{ n }{ { k }^{ 2 }+2k } =\displaystyle\sum _{ k=1 }^{ n }{ { k }^{ 2 } } +2k\displaystyle\sum _{ k=1 }^{ n }{ k }$

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

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

$=\dfrac { n\left( n+1 \right) \left( 2n+7 \right) }{ 6 }$

For any

$+$ ve integer

$n, n^3 + 2n$ is always divisible by

0%
$3$
0%
$7$
0%
$5$
0%
$6$

Explanation

Let

$P(n)=n^3+2n$

Now

$P(1)=1^3+2(1)=3$ , divisible by

$3$

$P(2)=2^3+2(4)=12$ , divisible by

$3$ and

$6$

$P(3)=3^3+2(3)=33$ , divisible by

$3$

Hence

$n^3+2n$ is divisible by

$3$

The last digit in

$\displaystyle { 7 }^{ 300 }$ is:

0%
7
0%
9
0%
1
0%
3

Explanation

We go on writing the units digit in the first few terms for the expansion of

$7$ .

$7^1 \rightarrow 7$

$7^2 \rightarrow 9$

$7^3 \rightarrow 3$

$7^4 \rightarrow 1$

$7^5 \rightarrow 7$

And the terms will keep on repeating.

Thus, the trend repeats itself in multiples of

$4$ .

Since

$300$ is divisible by

$4$ , the units digit of

$7^{300}$ would be

$1.$

Let

$A = \begin{pmatrix}1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\end{pmatrix}$ . Then for positive integer

$n, A^{n}$ is

0%
$\begin{pmatrix}1 & n & n^{2}\\ 0 & n^{2} & n\\ 0 & 0 & n\end{pmatrix}$
0%
$\begin{pmatrix}1 & n & n\left (\dfrac {n + 1}{2}\right )\\ 0 & 1 & n\\ 0 & 0 & 1\end{pmatrix}$
0%
$\begin{pmatrix}1 & n^{2} & n\\ 0 & n & n^{2}\\ 0 & 0 & n^{2}\end{pmatrix}$
0%
$\begin{pmatrix}1 & n & 2n - 1\\ 0 & \dfrac {n + 1}{2} & n^{2}\\ 0 & 0 & \dfrac {n + 1}{2}\end{pmatrix}$

Explanation

$A= \begin{pmatrix} 1& 1& 1\\ 0& 1& 1\\ 0& 0&1 \end{pmatrix}$

$A^{2}= \begin{pmatrix} 1& 2&3 \\ 0& 1&2 \\ 0& 0&1 \end{pmatrix}$

$A^{3}= \begin{pmatrix} 1& 3&6 \\ 0& 1&3 \\ 0& 0&1 \end{pmatrix}$
There using mathematical induction we get that

$A^{n}= \begin{pmatrix} 1& n& \frac{n(n+1)}{2} \\ 0& 1&n \\ 0& 0&1 \end{pmatrix}$

The number

$4^n \, + \, 15n \,-\, 1$ is a multiple of

$9$ for any natural

$n$ .

0%
True
0%
False

Explanation

$P(n):{ 4 }^{ n }+15n-1$ is a multiple of 9 for any natural number n.

$4^{n}$ can be written as

$(1+3)^{n}$ whose binomial expansion gives :

$1+3n+9(^{ n }{ C }_{ 2 })+........+{ 3 }^{ n }+15n-1$

which reduces to :

$18n+9(^{ n }{ C }_{ 2 })+........+{ 3 }^{ n }=9k$

where k is an integer.

Therefore,

${ 4 }^{ n }+15n-1$ is divisible by 9.

State true or false

$1 +$

$\dfrac{x}{a_1} \, + \, \dfrac{x(x \, + \, a_1)}{a_1 \, a_2} \, + \, ... \, + \, \dfrac{x(x \, + \, a_1) \, (x \, + \, a_2) \, .... \, (x \, + \, a_{n \, - \, 1})}{a_1 \, a_2 \, a_3 \,... \, a_n}$

$= \, \dfrac{(x \, + \, a_1) \, (x \, + \, a_2) \, .... \, (x \, + \, a_n)}{a_1 \, a_2 \, a_3 \,... \, a_n}$

0%
True
0%
False

Using the principle of mathematical induction ,

$\forall \, n\epsilon \, N$ if

$y = \cot$

$^{-1} \,$ x then

$y_n \, = \, (-1)^n\, (n \, - \, 1)!\, \sin^n$

$y \sin ny$ .

0%
True
0%
False

Prove by mathematical induction that
n. 1 + ( n - 1) 2 + (n - 2) 3 + ..... + 2 (n -1) + 1.n =

$\dfrac{n}{6}$ (n + 1)(n + 2)

0%
True
0%
False

Explanation

Result is obviously true for

$n \, = \, 1$ Assume P(n) i.e.

$P(n)\, = \, n\cdot 1\, + \, (n-1)\cdot 2\, + \, ....+ \, 2(n-1) \, + \, + \, 1\cdot n$

$= \, \dfrac{n}{6}\, (n+1)\,(n+2)$
p(n + 1) = (n + 1) 1 + n 2+ ,... 3(n -1)+2n + 1(n +1)

$\therefore \, \, p( n + 1) \, - \, p(n) \, = \, 1 \cdot 1 \, + \, 1 \, \cdot \, 2 \, + \, 1 \, \cdot \, 3 \, + \, 1 (n \, - \, 1 ) + \, 1 \, \cdot \, n \, + \, 1 \, + 1 \, \cdot \, (n \, + \, 1)$
= 1 + 2 + 3 ... + n + 1(n + 1)

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

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

$\dfrac{(n \, + \, 1)(n \, + \, 2) }{2} \, + \, \dfrac{(n \, + \, 1)(n \, + \, 2) }{6}$

$\dfrac{(n \, + \, 1)(n \, + \, 2) }{2} \, 1 \, + \, \left [ \dfrac{n}{3}\right]$

$\dfrac{(n \, + \, 1)(n \, + \, 2) (n \, + \, 3)}{6} \, = \, \dfrac{n}{6} \, (n \, + \, 1)(n \, + 2)$
Thus p (n + 1) also holds good

State the whether the statement id True/False

The given relation is

${ \left( 1+x \right) }^{ n }\ge \left( 1+nx \right)$ if

$x\ge -1$

0%
True
0%
False

If $A=\left ( \begin{matrix}cos \theta & isin \theta \\isin \theta & cos \theta\end{matrix} \right )$ , where $i = \sqrt { - 1}$ , then by principle of Mathematical Induction then $A^n=\left [ \begin{matrix}cos\,n \theta & isin\,n \theta \\isin\,n \theta & cos\,n \theta\end{matrix} \right ]$ .

0%
True
0%
False

Explanation

Given:

$A=\left[\begin{matrix} \cos{\theta} & i\sin{\theta} \\ i\sin{\theta} & \cos{\theta} \end{matrix} \right]$

To Check whether the given statement is true

${A}^{n}=\left[\begin{matrix} \cos{n\theta} & i\sin{n\theta} \\ i\sin{n\theta} & \cos{n\theta} \end{matrix} \right]$

Put

$n=1$

${A}^{1}=\left[\begin{matrix} \cos{\theta} & i\sin{\theta} \\ i\sin{\theta} & \cos{\theta} \end{matrix} \right]$

So,

${A}^{n}$ is true for

$n=1$

Let

${A}^{n}$ is true for

$n=k$ , so

${A}^{k}=\left[\begin{matrix} \cos{k\theta} & i\sin{k\theta} \\ i\sin{k\theta} & \cos{k\theta} \end{matrix} \right]$ ......

$(1)$

${A}^{k+1}=\left[\begin{matrix} \cos{\left(k+1\right)\theta} & i\sin{\left(k+1\right)\theta} \\ i\sin{\left(k+1\right)\theta} & \cos{\left(k+1\right)\theta} \end{matrix} \right]$

Now,

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

$=\left[\begin{matrix} \cos{k\theta} & i\sin{k\theta} \\ i\sin{k\theta} & \cos{k\theta} \end{matrix} \right]\left[\begin{matrix} \cos{\theta} & i\sin{\theta} \\ i\sin{\theta} & \cos{\theta} \end{matrix} \right]$

$=\left[\begin{matrix} \cos{k\theta}\cos{\theta}+{i}^{2}\sin{k\theta}\sin{\theta} & \cos{k\theta}\sin{\theta}+i\sin{k\theta}\cos{\theta} \\ i\sin{k\theta}\cos{\theta}+i\cos{k\theta}\sin{\theta} & {i}^{2}\sin{k\theta}\sin{\theta}+\cos{\theta}\cos{k\theta} \end{matrix} \right]$

$=\left[\begin{matrix} \cos{k\theta}\cos{\theta}-\sin{k\theta}\sin{\theta} & i\sin{k\theta}\cos{\theta}+\cos{k\theta}\sin{\theta} \\ i\left(\sin{k\theta}\cos{\theta}+\cos{k\theta}\sin{\theta}\right) & \cos{\theta}\cos{k\theta}-\sin{k\theta}\sin{\theta} \end{matrix} \right]$

$=\left[\begin{matrix} \cos{\left(k+1\right)\theta} & i\sin{\left(k+1\right)\theta} \\ i\sin{\left(k+1\right)\theta} & \cos{\left(k+1\right)\theta} \end{matrix} \right]$

So,

${A}^{n}$ is true for

$n=k+1$ whenever it is true for

$n=k$

Hence, by principle of mathematical induction

${A}^{n}$ is true for all positive integer.

Hence the given statement is true.

State whether the statement is true/false

$a$ and

$b$ are

$2$ +ve no. such that

$a>b$ then one of the two numbers

$\dfrac { a+b }{ 2 }$ and

$\dfrac { a-b }{ 2 }$ is even or odd

0%
True
0%
False

State true or false.

${1^3}\, + \,{3^3}\, + \,{5^3}\, + \,...\, + \,{\left( {2n\, - \,1} \right)^3}\, = \,{n^2}\,\left( {2{n^2}\, - \,1} \right)$ n is a natural number

0%
True
0%
False

Explanation

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

The result is true for

$n= 1$

$2{n}^{4}-{n}^{2}=2{\left(1\right)}^{4}-{\left(1\right)}^{2}=2-1=1$

Let the result be true for

$n = k$ . That is

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

Now we need to prove that the result is also true for

$n = k+ 1$ . That is

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

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

$=2{k}^{4}-{k}^{2}+8{k}^{3}+3\times 4{k}^{2}+3\times 2k+1\\$

$=2{k}^{4}+8{k}^{3}+12{k}^{2}+8k+2-2k-1-{k}^{2}\\$

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

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

$\therefore\,$ The result is also true for

$n = k + 1$ .

Hence by the principle of mathematical induction the result is true for all

$n\in N$

State true or false

$1.3\, + \,{\left( {{{2.3}^{}}} \right)^2}\, + \,{3.3^3}\, + \,.....\, + \,n{.3^n}\, = \,\dfrac{{\left( {2n\, - \,1} \right){3^{n\, + \,1}}\, + \,3}}{4}$

0%
True
0%
False

State whether following statement is true or false.

By using the principle of mathematical induction we can proove that

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

$3$ .

0%
True
0%
False

Explanation

Let

$p(n)=n(n+1)(n+5)=3d$

where

$d \epsilon N$

For

$n=1$

$LHS=1(1+1)(1+5)$

$=1(2)(6)$

$=12$

$RHS$

$=3 d=3\times 4=12$

$\therefore p(n)$ is true for

$n=1$

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

$3$ .

For every positive integer

$n$ ,

${7^n} - {3^n}$ is divisible by

$4$ .

0%
True
0%
False

Explanation

Consider

$P(n):{7^n} - {3^n}$ is divisible by

$4$

Now,

$P(1):{7^1} - {3^1} = 4$

Thus, it is true for

$n=1$

Let

$p(k)$ is true for

$n=K$

${7^k} - {3^k}\,is\,divisible\,by\,4$

Now, prov that

$P(k+1)$ is true.

$\begin{array}{c} { 7^{ (k+1) } }-{ 3^{ (k+1) } }={ 7^{ (k+1) } }-{ 7.3^{ k } }+{ 7.3^{ k } }-{ 3^{ (k+1) } } \\ \qquad \qquad \: \: \: =7({ 7^{ k } }-{ 3^{ k } })+(7-3){ 3^{ k } }=7(4d)+(7-3){ 3^{ k } } \\ \qquad \qquad \: \: \: =7(4d)+{ 4.3^{ k } }=4(7d+{ 3^{ k } }) \end{array}$

Hence,

$P(n):{7^n} - {3^n}$ is divisible by

$4$ is true.

$1.3 + {\left( {2.3} \right)^2} + {(3.3)^3} + .... + {(n.3)^n} = \frac{{\left( {2n - 1} \right){3^{n + 1}} + 3}}{4}$

0%
True
0%
False

Explanation

To prove or disprove

$1.3 + (2.3)^2 + (3.3)^3 + ..... + (n.3)^n = \dfrac{(2n-1)3^{n+1}+3}{4}$

$P(n)=1.3 + (2.3)^2 + (3.3)^3 + ...... + (n.3)^n = \dfrac{(2n-1)3^{n+1}+3}{4}$

Let

$n = 1$ ,

$P(1) = 1.3 = \dfrac{(2\times 1 - 1)3^{1+1} + 3}{4}$

$\because 1.3 = 3 = LHS$

$\dfrac{1\times 3^2 + 3}{4} = \dfrac{9+3}{4} = \dfrac{12}{4} = 3 = RHS$

$\therefore LHS = RHS$

$P(n)$ is true for

$n = 1$

Assume

$P(k)$ is true,

i.e,

$(1.3) + (2.3)^2 + (3.3)^3 + ....... + (k.3)^k = \dfrac{(2k-1)3^{k+1}+3}{4}$ ...(1)

Now let us prove that

$P(k+1)$ is true,

LHS:

$1.3 + (2.3)^2 + (3.3)^3 + ... (k.3)^k + [(k+1).3]^{k+1}$

substitute eq (1),

$= \dfrac{(2k-1)3^{k+1}+3}{4} + [(k+1) .3]^{k+1}$

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

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

$= \dfrac{(2k+1)3^{k+1+1}+3}{4} = \frac{[2(k+1)-1]3^{(k+1)+1}+3}{4}$

RHS:

$\dfrac{[2(k+1)-1]3^{(k+1)+1}+3}{4}$

LHS = RHS when

$P(k)$ is true.

hence proved.

State True or False

${7^n} - {2^n}$ is divisible by

$5$ .

0%
True
0%
False

${5}^{2n+2}-24n-25$ is divisible by

$576$ for all

$n\in N$ by using principle of mathematical induction.

0%
True
0%
False

State whether the following statement is true or false.

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

0%
True
0%
False

Using the principle of mathematical induction for all

$n\in N$ it can be proved

$1+2+3+...+n < \dfrac{1}{8}(2n+1)^2$ .

0%
True
0%
False

CBSE Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.