If $$a_1 = 1, a_{n+1} = \dfrac{1}{n+1}a_n,\forall n \geq 1$$, then $$a_n \ =$$

$$\displaystyle \frac{1}{n !}$$

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

$$\displaystyle \frac{1}{(n + 1)!}$$

MCQ Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 6

$n$ is an odd positive integer, then

$a^n + b^n$ is divisible by

0%
$a - b$
0%
$a + b$
0%
$a^2 + b^2$
0%
none of these

Explanation

Let,

$P(n) =a^n+b^n$

$P(1) = a+b$ , which is divisible by

$a+b$

Now let

$P(k) =a^k+b^k$ is divisible by

$a+b$ , where

$k$ is an odd integer.

$\Rightarrow a^k+b^k = (a+b)f(a,b)......................... (1)$

Now,

$P(k+2)= a^{k+2}+b^{k+2} =a^2[(a+b)f(a,b)-b^k]+b^{k+2}$

$\quad \quad =a^2f(a,b)(a+b)-a^2b^k+b^{k+2}$ from

$(1)$

$\quad \quad =a^2f(a,b)(a+b)-b^k(a^2-b^2)$

$\quad \quad =(a+b)\left[a^2f(a,b)-b^k(a-b)\right]$ , which is divisible by

$(a+b)$

Hence

$a^n+b^n$ is divisible by

$(a+b)$ for all odd positive integral

$n$

The value of

$\displaystyle \frac{1^2}{1.3} + \frac{2^2}{3 . 5}+\dots+ \frac{n^2}{(2n - 1)(2 n + 1)}$ is

0%
$\displaystyle \frac{n (n + 1)}{2 (2n + 1)}$
0%
$\displaystyle \frac{n (n - 1)}{2 (2n - 1)}$
0%
$\displaystyle \frac{n^2 (n - 1)^2}{2 (2n + 1)}$
0%
none of these

Explanation

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

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

$=\dfrac{1}{4}[\dfrac{4n^{2}-1+1}{4n^{2}-1}]$

$=\dfrac{1}{4}[1+\dfrac{1}{4n^{2}-1}]$

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

$=\dfrac{1}{4}[1+\dfrac{1}{2}(\dfrac{1}{2n-1}-\dfrac{1}{2n+1})]$

Consider

$\dfrac{1}{2n-1}-\dfrac{1}{2n+1}$

$=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}...\dfrac{1}{2n-1}-\dfrac{1}{2n+1}$

$=1-\dfrac{1}{2n+1}$

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

Hence applying summation gives us

$\dfrac{n}{4}+\dfrac{1}{8}\sum (\dfrac{1}{2n-1}-\dfrac{1}{2n+1})$

$=\dfrac{n}{4}+\dfrac{2n}{8(2n+1)}$

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

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

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

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

The value of

$\displaystyle \frac{1}{1.2.3} + \frac{1}{2.3.4}+ ....+ \frac{1}{n (n + 1) (n + 2)}$ is

0%
$\displaystyle \frac{n(n + 3)}{4(n + 1) (n + 2)}$
0%
$\displaystyle \frac{n}{(n + 1)(n + 2)}$
0%
$\displaystyle \frac{n (n + 2)}{ (n + 1)(n + 3)}$
0%
All of these

Explanation

Let

$P(n) : \displaystyle \frac{1}{a .2. 3} + \frac{1}{2.3.4} + .....+ \frac{1}{n(n + 1) (n+2)} = \frac{n(n+3)}{4(n + 1) (n + 2)}$ ...... (1)

Step I: For

$n = 1$ ,

L.H.S. of

$(1) = \displaystyle \frac{1}{1.2.3} = \frac{1}{6}$

and R.H.S of

$(1) = \displaystyle \frac{1. (1 + 3)}{4(1 +1)(1+2)} = \frac{1}{6}$

Therefore P(1) is true.

Step II : Assume that P(k) is true, then

$\displaystyle P(k) : \frac{1}{1.2.3} + \frac{1}{2.3.4} + ....+ \frac{1}{k(k + 1)(k + 2)} = \frac{k(k + 3)}{4(k + 1)(k+2)}$

Step III : For

$n = k +1$ ,

$\displaystyle P(k + 1) : \frac{1}{1.2.3} + \frac{1}{2.3.4} + .... + \frac{1}{k(k + 1)(k+2)} + \frac{1}{(k + 1)(k+2)(k+3)}$

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

$\therefore$ L.H.S

$= \displaystyle \frac{1}{1.2.3} + \frac{1}{2.3.4} + ..... + \frac{1}{k(k +1) (k + 2)} + \frac{1}{(k +1)(k + 2)(k+3)}$

$= \displaystyle \frac{k(k+3)}{4(k + 1)(k+2)} + \frac{1}{(k + 1)(k+2)(k + 3)}$ (By assumption step)

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

$=\displaystyle \frac{k^3 + 6k^2 + 9k + 4}{4(k + 1)(k +2)(k+3)}$

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

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

$= R. H. S$

Hence P(k + 1) is true. Hence by the principle of mathematical induction P(n) is true for all n

$\in$ N.

The value of

$\displaystyle \tan^{-1} \left ( \frac{1}{3} \right ) + \tan^{-1} \left ( \frac{1}{7} \right ) +\dots+ \tan^{-1} \left ( \frac{1}{n^2 + n + 1} \right )$ is

0%
$\displaystyle \tan^{-1} \left ( \frac{n}{n + 2} \right )$
0%
$\displaystyle \tan^{-1} \left ( \frac{n+1}{n - 1} \right )$
0%
$\displaystyle \tan^{-1} \left ( \frac{n-1}{n + 2} \right )$
0%
$\displaystyle \tan^{-1} \left ( \frac{n+2}{n - 2} \right )$

Explanation

$T_n = \tan^{-1}\left(\cfrac{1}{1+n(1+n)}\right)= \tan^{-1}\left(\cfrac{(n+1)-n}{1+n(1+n)}\right)=\tan^{-1}(n+1)-\tan^{-1}n$

Hence required sum is

$=\sum T_n$

$=[\tan^{-1}(2)-\tan^{-1}1]+[\tan^{-1}(3)-\tan^{-1}2]+........+\tan^{-1}(n)-\tan^{-1}(n-1)]+[\tan^{-1}(n+1)-\tan^{-1}n]$

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

By mathematical induction

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

0%
$p^2 + p + 1$
0%
$p^2 + 1$
0%
$p+1$
0%
None of these

Explanation

Let

$f\left( n \right)={ p }^{ n+1 }+{ \left( p+1 \right) }^{ 2n-1 }$

we have

$f\left( 1 \right) ={ p }^{ 2 }+p+1$ which is divisible by

${p}^{2}+p+1$

Now, assume that

$f(m)$ is divisible by

${p}^{2}+p+1$

$\therefore{ p }^{ m+1 }+{ \left( p+1 \right) }^{ 2m-1 }=k\left( { p }^{ 2 }+p+1 \right)$ ...(1)

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

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

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

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

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

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

Hence,

$f(m+1)$ is divisible by

${p}^{2}+p+1$

Hence by mathematical induction

$f(n)$ is divisible by

${p}^{2}+p+1$ for all

$n\in N$

$a_1 = 1, a_{n+1} = \dfrac{1}{n+1}a_n,\forall n \geq 1$ , then

$a_n \ =$

0%
$\displaystyle \frac{1}{n !}$
0%
$\displaystyle \frac{1}{(n + 2)!}$
0%
$\displaystyle \frac{1}{(n + 1)!}$
0%
none of these

Explanation

$a_1=1, a_2 = \cfrac{1}{1+1}a_{1}=\cfrac{1}{2}(1)=\cfrac{1}{2!}, a_3 = \cfrac{1}{3}.\cfrac{1}{2!} = \cfrac{1}{3!}$
Similarly

$a_n = \cfrac{1}{n}.\cfrac{1}{(n-1)!} = \cfrac{1}{n!}$

$3 + 13 + 29 + 51 + 79 + ...$ to

$n$ terms

$=$

0%
$2n^2 + 7n^3$
0%
$n^2 + 5n^3$
0%
$n^3 + 2n^2$
0%
none of these

Explanation

Let

$S_{n}=3+13+29+51+79+\cdots +n terms$

$\because S_{1}=3$ and

$S_{2}=3+13=16$

Similarly

$S_{3}=3+13+29=45$

By options, from

$(c)$ as we have

$S_{n}=n^{3}+2n^{2}$

check by putting

$n=1, S_{1}=1^{3}+2.1^{2}=3$

$S_{2}=2^{3}+2.2^{2}=16$ and

$S_{3}=3^{3}+2.3^{2}=45$

Thus option

$(c)$ is correct

Using the principle of mathematical induction, find

$tan \alpha + 2 tan 2 \alpha + 2^2 tan 2^2 \alpha + ....$ to

$n$ terms:

0%
$tan \alpha - 2^n \cdot tan (2^n \cdot \alpha)$
0%
$cot \alpha - 2^n \cdot cot (2^n \cdot \alpha)$
0%
$sec \alpha - 2^n \cdot sec (2^n \cdot \alpha)$
0%
None of these

Explanation

Let P(n):

$tan \alpha + 2 tan 2\alpha + 2^2 tan 2^2 \alpha + .... + 2^{n - 1} tan (2^{n -1} \alpha) = cot \alpha - 2^n \cdot cot (2^n \cdot \alpha)$ ..... (1)

Step I: For

$n = 1$ ,

L.H.S. of

$(1) = tan \alpha$

$= cot \alpha - cot \alpha + tan \alpha = cot \alpha- (cot \alpha - tan \alpha)$

$= cot \alpha- \displaystyle \left ( cot \alpha - \frac{1}{cot \alpha} \right )$

$= cot \alpha - 2 \left ( \dfrac{cot^2 \alpha - 1}{2 cot \alpha} \right )$

$= cot \alpha - 2 \left ( \dfrac{cos^2 \alpha - \sin ^2 \alpha}{2 cos \alpha \sin \alpha} \right )$

$= cot \alpha - 2 cot 2 \alpha$

$= R. H.S.$ of (1)

Therefore,

$P(1)$ is true.

Step II: Assume it is true for

$n = k$ , then

$P(k) : tan \alpha + 2 tan 2\alpha + 2^2 tan 2^2\alpha + .... + 2^{k - 1} tan (2^{k - 1} \alpha)$

$= cot \alpha - 2^k cot (2^k \alpha)$

Step III: For

$n = k + 1$ ,

$P(k +1) : tan \alpha + 2 tan 2 \alpha + 2^2 tan 2^2 \alpha + ... + 2^{k - 1} tan (2^{k - 1} \alpha) + 2^k tan (2^k \alpha) = cot \alpha - 2^{k + 1} cot (2^{k + 1} \alpha)$

$L.H.S. = tan \alpha + 2 tan 2\alpha + 2^2 tan 2^2 \alpha + .... + 2^{k - 1} tan (2^{k - 1} \alpha) + 2^k tan (2^k \alpha)$

$= cot \alpha - 2^k cot (2^k \alpha) + 2^k tan (2^k \alpha)$ (By assumption step)

$= cot \alpha - 2^k (cot (2^k \alpha) - tan (2^k \alpha))$

$= cot \alpha - 2^k \cdot 2 \displaystyle \left ( \frac{cot^2 (2^k \alpha) - 1}{2 cot (2^k \alpha)} \right )$

$= cot \alpha - 2^{k + 1} \cdot cot (2 \cdot 2^k \alpha)$

$= cot \alpha - 2^{k + 1} \cdot cot (2^{k + 1} \alpha)$

$= R.H.S.$

This show that the result is true for

$n = k + 1$ .

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

$\in$ N.

$p$ is a prime number, then

$n^p -n$ is divisible by

$p$ for all

$n$ , where

0%
$n\in N$ .
0%
$n$ is odd natural number.
0%
$n$ is even natural number.
0%
$n$ is not a composite number.

Explanation

$n^{p}-n$ is divisible by

$p$ for

$n=1$
Let,

$k^{p}-k=p\lambda$ , where

$k\in N$ and

$\lambda \in Z$

Now,

$\left (k+1 \right )^{p}-\left (k+1 \right )=k^{p}+^{p}\textrm{C}_{1}k^{p- 1}+.........+^{p}\textrm{C}_{p}-1-k$

$\left (k^{p}-k \right )+\left (^{p}\textrm{C}_{1}k^{p-1}+...+^{p}\textrm{C}_{p-1}k \right )$

$=p\lambda +\left (^{p}\textrm{C}_{1}k^{p-1}+...+^{p}\textrm{C}_{p-1}k \right )$
which is divisible by

$p$
Reason :

$^{p}\textrm{C}_f$ is always divisible by by p when

$f\neq p,0$ because p is a prime number as the denominator will not have any factors or multiples of p.

So by mathematical induction,

$n^{p}-n$ is divisible by

$p$ for all

$n\in N$

For each

$n\, \epsilon\, N$ , then

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

0%
$2$
0%
$3$
0%
$7$
0%
None of these

Explanation

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

Let

$n=1$ .

Then,

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

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

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

${ 3 }^{ 3 }+1=28$ which is divisible by

$2.$

Let

$n=2$ .

Then,

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

${ 3 }^{ 5 }+1$

$243+1=244$ which is divisible by

$2.$

For positive integer n,

$3^n < n!$ when

0%
$n\geq 6$
0%
$n > 7$
0%
$n\geq 7$
0%
$n \leq 7$

Explanation

For positive integer

$n$ ,

${ 3 }^{ n }<n!$

Let

$n=1$

${ 3 }^{ 1 }<1!$

$={ 3 }<1$ which is not possible.

Let

$n=2$

${ 3 }^{ 2 }<2!$

$={ 9 }<2$ which is not possible.

Let

$n=6$

${ 3 }^{6 }<6!$

$={729 }<720$ which is not possible.

Let

$n=7$

${ 3 }^{7 }<7!$

$={2187 }<5040$ which is possible.

$\therefore$ For positive integer n,

${ 3 }^{ n }<n!$ when

$n\ge 7$

$\forall$

$n\in N$ ,

$\displaystyle 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ...... + \frac{1}{\sqrt{n}}$ is

0%
$\sqrt{n}$
0%
$\leq \sqrt{n}$
0%
$\gt\sqrt{n}$
0%
none of these

Explanation

Let

$P(n)=\displaystyle 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ...... + \frac{1}{\sqrt{n}}$ ....(1)

Put

$n=2$ in (1)

$P(2)=\displaystyle 1 + \frac{1}{\sqrt{2}} > \sqrt{2}$

Put

$n=3$ in (1)

$P(3)=\displaystyle 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} >\sqrt{3}$

So, let us assume

$P(n)=\displaystyle 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ...... + \frac{1}{\sqrt{n}} >\sqrt{n}$

So, we will check for

$P(n+1)$

$P(n+1)= \displaystyle 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ...... + \frac{1}{\sqrt{n}} +\frac{1}{\sqrt{n+1}}$

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

Since,

$\sqrt{n(n+1)}>n$

$\Rightarrow \sqrt{n(n+1)}+1 >n+1$

$\Rightarrow \dfrac{\sqrt{n(n+1)}+1}{\sqrt{n+1}}>\dfrac{n+1}{\sqrt{n+1}}$

$\Rightarrow \dfrac{\sqrt{n(n+1)}+1}{\sqrt{n+1}}>\sqrt{n+1}$ ....(3)

From (2) and (3), it follows that

$\displaystyle 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ...... + \frac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}>\sqrt{n+1}$

Hence, by mathematical induction method, we can say that

$\displaystyle 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ...... + \frac{1}{\sqrt{n}}>\sqrt{n}$ for

$n\in N, n\ge 2$

Let

$\displaystyle p(n)=x\left(x^{n-1}-n\cdot a^{n-1}+a^{n}(n-1)\right)\>$ is divisible by

$\left ( x-a \right )^{2}$ for

0%
$n> 1$
0%
$n> 2$
0%
$\forall \;n\;\in \;N$
0%
None of these

Explanation

$p(1)=x$
Hence it is not divisible by

$(x-a)^{2}$ .

$p(2)=x(x-2a+a^{2})$

$=x^{2}-2ax+a^{2}x$

$=(x-a)^{2}-a^{2}+a^{2}(x)$

$=(x-a)^{2}+a^{2}(x-1)$ ...(i)
Hence

$p(2)$ is also not divisible by

$(x-a)^{2}$ .
Now

$\{1,2\}$ falls in the set of first n natural numbers.
Hence the correct answer is Option D.

$10^n + 3\cdot 4^{n +2} + \lambda$ is exactly divisible by

$9$ for all

$n\in N$ , then the least positive integral value of

$\lambda$ is

0%
$5$
0%
$3$
0%
$7$
0%
$1$

Explanation

Let

$P(n)=10^n + 3\cdot 4^{n +2} + \lambda$
Since

$P(n)$ is divisible by

$9$ for all

$n\in N$

$\Rightarrow P(1)$ will be also divisible by 9

$\Rightarrow P(1)=10+3(4^3)+\lambda=202+\lambda=22\times 9+(\lambda+4)$
Thus least positive integral value of

$\lambda$ for which

$P(1)$ can be divisible by

$9$ is

$5$
Hence option 'A" is correct choice.

For all

$n \in N$ ,

$10^n + 3.4^{n+2} + 5$ is divisible by

0%
$23$
0%
$3$
0%
$9$
0%
$207$

Explanation

$10^n + 3.4^{n+2} + 5$
Put

$n=1$

$10+3(64)+5=207$ which is divisible by

$3,9,23,207$

We will check for

$n=2$

$100+768+5=873$
which is divisible by

$3$ and

$9$ only.

Every number divisible by

$9$ is also divisible by

$3$ .
Now, let

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

$10^n + 3.4^{n+2} + 5$ is divisible by

$9$

$\Rightarrow 10^n + 3.4^{n+2} + 5=9\lambda$ ....(1)

Now, we will check for

$P(n+1)$
Consider,

$10^{n+1} + 3.4^{n+3} + 5$

$=10^n10+3.4^{n+2}4+5$

$=10(9\lambda -3.4^{n+2}-5)+3.4^{n+2}4+5$ (by (1))

$=90\lambda-27.4^{n+2}-45$

$=9(10\lambda-3.4^{n+2}-5)$
which is divisible by 9.

Hence, by mathematical induction given expression is divisible by

$3,9$ .

$^{4n}{C_{{\text{2n}}}}{:^{2n}}{{\text{C}}_{\text{n}}} = \{ 1.3,.5,...(4{\text{n}} - 1)\} :{\{ 1,3,5....(2{\text{n}} - 1)\} ^\lambda },then\;\lambda =$

0%
1
0%
2
0%
3
0%
none of these

Explanation

$^{4n}C_{2n}:^{2n}C{n}=\left\{1,3,5,….\left(4n-1\right)\right\};\left\{1,3,5,…. \left(2n-1\right)\right\}$

We have

$\dfrac{2n!}{n!}=\dfrac{2n\left(2n-1\right) \left(2n-1\right)….3.2.1}{n!}$

$=\left[2.4.6….\left(2n-2\right) \left(2n\right)\right]\left[1.3.5….\left(2n-1\right)\right]/n!$

Writing the odd terms and even terms we get

$\dfrac{2n!}{n!} = 2^{n}\left[1.2.3….\left(n-1\right).n\right] \left[1.3.5….\left(2n-1\right)\right]/n!$

$=\dfrac{2^{n}n! \left[1.3.5….\left(2n-1\right).n\right]}{n!}= \left[1.3.5….\left(2n-1\right)\right]$

Replacing

$n$ by

$2n$ ,

$\dfrac{4n!}{2n!}=2^{2n}\left[1.3.5….\left(4n-1\right)\right]$

Now we have

$^{4n}C_{2n}:^{2n}C_{n}=\dfrac{4n!}{2n!2n!}\times\dfrac{n!n!}{2n!}=\dfrac{4n!}{2n!}.\left(\dfrac{n!}{2n!}\right)^{2}$

Now

$^{4n}C_{2n}!^{2n}C_{n}=2^{2n}\dfrac{\left[1.3.5….\left(4n-1\right)\right]}{\left[2^{n}\left(1.3.5….(2n-1)\right)\right]}$

$= \dfrac{\left[1.3.5….\left(4n-1\right)\right]}{\left[ 1.3.5….\left(2n-1\right)\right]^{2}} \Rightarrow \lambda=2$

$2^{83}+k$ is divisible by 127, then the smallest positive integral value of k is:

0%
63
0%
31
0%
15
0%
64

Using principle of mathematics induct or for all

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

0%
True
0%
False

For all

$n \in N ,$

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

0%
True
0%
False

$\displaystyle{\frac { 1 }{ { 1 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 2 } } +\frac { 3 }{ { 3 }^{ 2 } } +......+\frac { 1 }{ { n }^{ 2 } } \le \frac { 3n+1 }{ 2n+2 } }$ for every natural number

$n$

0%
True
0%
False

For any natural number

$n,$

$x^n y^n$ is divisible by

$x y,$ where

$x$ and

$y$ are any integers with

$x \neq y.$

0%
True
0%
False

CBSE Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 6 - 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 6 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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