NCERT Exemplar Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction
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NCERT Exemplar Class 11 Maths Solutions Chapter 4 Principle of Mathematical Induction

Komal MiglaniUpdated on 31 Mar 2025, 01:16 AM IST

You've always wondered how to prove the truth of a pronouncement for all natural numbers or how to prove that a form embraces indefinitely. The answer lies in the principle of Mathematical Induction, which is explained in Class 11, Chapter 4 of mathematics. Mathematical induction is a technique to prove mathematical results that hold for all natural numbers. This article gives you an introduction to the principle of induction and helps you to understand how it is used as a fundamental tool in mathematics. You will understand how to find the truth of a mathematical statement that seems to be complicated beyond the initial, but can still be proved easily by applying induction. Your ability to solve difficult problems will be greatly improved if you understand the method of inductive proofs, where you prove a result for some values and then assume that the statement remains unchanged for other sets of values.

Regular practice with NCERT Exercises and NCERT Worksheets will help you achieve certainty and sharpen your logical deduction skills, preparing you for joint examinations and practical troubleshooting. Hence, we must be ready to unlock the powers of mathematical initiation and to discover the way in which the tenet can connect and simplify the apparently complex mathematical notions.

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  1. NCERT Exemplar Class 11 Maths Solutions Chapter 4
  2. Subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 4
  3. NCERT Solutions for Class 11 Maths: Chapter Wise
  4. Importance of solving NCERT Exemplar Class 11 Maths Questions
  5. NCERT Exemplar Class 11 Mathematics Chapters
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NCERT Exemplar Class 11 Maths Solutions Chapter 4

Class 11 Maths Chapter 4 exemplar solutions Exercise: 4.3
Page number: 70-72
Total questions: 30

Question:1

Give an example of a statement $P(n)$ which it true for all $n \geq 4$ but $P(1), P(2) \: \: and \: \: P(3)$ are not true. Justify your answer.

Answer:

Given:
$P(n)$ which it true for all $n \geq 4$ but $P(1), P(2) \: \: and \: \: P(3)$ are not true.

Let us consider P (n) be $2^n< n!$

Examples of the given statements are-

$P (0) = 2^0<0!$ i.e., $1<1 \rightarrow Not True$

$P (1) = 2^1<1!$ i.e., $2<1 \rightarrow Not True$

$P(2) = 2^2<2!$ i.e., $4<2 \rightarrow$ NotTrue
$P(3) = 2^3<3!$ i.e., $8<6 \rightarrow$ NotTrue
$P(4) = 2^4<4!$ i.e., $16<24 \rightarrow$ True
$\& P(5) = 2^5<5!$ i.e., $32<60 \rightarrow$ True.....&so on.

Question:2

Give an example of a statement P (n) which is true for all n. Justify your answer.

Answer:

P (n) is true for all n …….. (Given)

Now, let P (n) be,

$1+ 2 + 3 + 4 +... + n = n (n+1)/2$
P(0) $=0(0+1)/2 = 0 $

$\rightarrow true\\ P (1) $

$\rightarrow1 = 1(2+2)/2 = 1 $

$\rightarrow true\\ P (2)$

$ \rightarrow 1+2 = 2(2+1)/2 $

$\rightarrow true\\ P (k) $

$\rightarrow 1+2+3+... +k = k (k+1)/2\\ P (k+1) $

$\rightarrow 1+2+3+... +k+1 = k (k+1)/2 + k + 1 = (k+1) (k+2)/2 \rightarrow p (k+1)$

is true for all k

Thus, P (n) is true for all n.

Question:3

Prove each of the statements in Exercises 3 to 16 by the Principle of Mathematical Induction.
$4^n -1$ is divisible by 3, for each natural number n.

Answer:

P (n) = $4^n -1$ is divisible by 3 …….. (Given)

Now, we’ll substitute different values for n,

$P (0) = 4^0 - 1 = 0$, viz. divisible by 3

$P (1) = 4^1 - 1 = 3$, divisible by 3.

$P (2) = 4^2 - 1 = 15$, divisible by 3

Now, let $P (k) = 4^k - 1$, is divisible by 3

Thus, $4^k - 1 = 3x$

We also get that,

$P (k+1) = 4^{k+1} - 1 = 4(3x+1) - 1$

$= 12x + 3$, is divisible by 3

Thus, P (k+1) is also true,

Hence, by mathematical induction,

For each natural number n it is true that, $P (n) = 4n -1$ is divisible by 3

Question:4

$2^{3n} -1$ is divisible by7, for all natural numbers n.

Answer:

Now, we'll substitute different values for n,

$
\begin{aligned}
& \mathrm{P}(0)=2^0-1=0, \text { divisible by } 7 \\
& P(1)=2^3-1=7, \text { divisible by } 7 \\
& P(2)=2^6-1=63 \text {, divisible by } 7 \\
& P(3)=2^9-1=512 \text {, divisibleby } 7
\end{aligned}
$
Now, let us consider,

$
P(k)=2^{3 k}-1 \text { be divisible by } 7
$
Thus, $2^{2 k}-1=7 x$
We also get that,

$
P(k+1)=2^{3(k+1)}-1=2^3(3 x+1)-1=56 x+7, \text { is divisible by } 7
$
Thus, $\mathrm{P}(\mathrm{k}+1)$ is also true,
Hence, by mathematical induction, For each natural number n it is true that,

$
\mathrm{P}(\mathrm{n})=2^{3 \mathrm{n}}-1 \text { is divisible by } 7.
$

Question:5

$n^3 -7n + 3$ is divisible by 3 for all natural numbers n.

Answer:

$n^3 -7n + 3$ is divisible by 3... given

Now, we’ll substitute different values for n,

$P (0) = 0^3 -7 \times 0 + 3 = 3$, is divisible by 3

$P (1) = 1^3 -7 \times 1 + 3 = -3$, is divisible by 3

$P (2) = 2^3 -7 \times 2 + 3 = -3$, is divisible by 3

Now, let us consider,

$P (k) = k^3 -7k + 3$ be divisible by 3

Thus, $k^3 -7k + 3 = 3x$

We also get that,

$P (k+1) = (k+1)^3 -7(k+1) + 3 = k^3 + 3k^2 + 3k +1 -7k -7 + 3$

$= 3x + 3(k^2+k-2)$is divisible by 3

Thus, P (k+1) is also true,

Hence, by mathematical induction,

For each natural no. n it is true that, $P (n) = n^3 -7n + 3$ is divisible by 3.

Question:6

$3^{2n -1}$ is divisible by 8, for all natural numbers n.

Answer:

p(n)= $3^{2n -1}$ is divisible by 8.......(given)

Now, we’ll substitute different values for n,

$p(0) = 3^0 -1 = 0$, is divisible by 8

$P(1) = 3^2 -1 = 8$, is divisible by 8

$P(2) = 3^4 -1 = 80$ , is divisible by 8

Now, let us consider,

$P(k) = 3^{2k} -1$ be divisible by 8

Thus, $P(k+1) = 3^{2(k+1)} -1 = 72x + 8$, is divisible by 8

Hence, by mathematical induction,
For each natural number n, it is true that $P(n) = 3^{2n} -1$ is divisible by 8.

Question:7

For any natural number n,$7^n -2^n$ is divisible by 5.

Answer:

P(n)=$7^n -2^n$ is divisible by 5. ...............(given)

Now, we will substitute different values for n,

$P(0) = 7^0 -2^0 = 0$, is divisible by 5

$P(1) = 7^1 -2^1 = 5$, is divisible by 5

$P(2) = 7^2 -2^2 = 45$, is divisible by 5

$P(3) = 7^3 -2^3 = 335$, is divisible by 5

Now, let us consider that,

$P(k) = 7^k -2^k$ be divisible by 5.

Thus, $7^k -2^k=5x$

We also get,

$P(k+1) = 7^{k+1} -2^{k+1} = (5+2)7^k -2(2)^k$

$= 5(7)^k + 2(5x)$, is divisible by 5

Thus, P(k+1) is true

Hence, by mathematical induction,
For each natural number n it is true that $P(n) = 7^n -2^n$ is divisible by 5.

Question:8

For any natural number n, $x^n -y^n$ is divisible by x -y, where x integers with$x \neq y$.

Answer:

P(n)= $x^n -y^n$ is divisible by x -y, where x integers with$x \neq y$. ........(given)
Now, we’ll substitute different values for n,

$P(0) = x^0 -y^0 = 0$, is divisible by x-y

$P(1) = x -y,$ is divisible by x-y

$P(2)=x^2-y^2=(x+y)(x-y)$, is divisible by $\mathrm{x}-\mathrm{y}$
Now, let us consider,
$P(k)=x^k-y^k$, is divisible by x - y
Thus, $x^k-y^k=a(x-y)$
We also get that

$
\begin{aligned}
& P(k+1)=x^{k+1}-y^{k+1} \\
& =x^k(x-y)+y\left(x^k-y^k\right) \\
& =x^k(x-y)+y a(x-y), \text { is divisible by } \mathrm{x}-\mathrm{y}
\end{aligned}
$
Thus, $P(k+1)$ is true
Hence, by mathematical induction,
For each natural no. n it is true that, $P(n)=x^n-y^n$ is divisible by $\mathrm{x}-\mathrm{y}, \mathrm{x}$ integers with $\mathrm{x} \neq \mathrm{y}$.

Question:9

$n^3 -n$ is divisible by 6, for each natural number n. $n \geq2$

Answer:

P(n)= $n^3 -n$ is divisible by 6

Now, we’ll substitute different values for n,

$P(2) = 2^3 -2 = 6$, is divisible by 6

Now, let us consider,

$P(k) = k^3 -k$ be divisible by 6

Thus, $k^3 -k = 6x$

We also get that,

$P(k+1) =(k+1)^3 -(k+1)$

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

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

$= 6x + 3k(k+1)$

$= 6x + 3 \times 2y$, $\because k(k+1) \text{ is even}$

$= 6x + 3 \times 2y$ is divisible by 6

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural number n, it is true that P(n)= $n^3 -n$ is divisible by 6.

Question:10

$n(n^2 + 5)$ is divisible by 6 for each natural number n.

Answer:

P(n)= $n(n^2 + 5)$ is divisible by 6 for each natural number n.

Now, we’ll substitute different values for n,

$P(1) = 1(1^2 + 5) = 6$, is divisible by 6

$P(2) = 2(2^{}2 + 5) = 18$, is divisible by 6

Let us consider,

$P(k) = k(k^2 + 5)$be divisible by 6

Thus, $k(k^2 + 5)=6x$

We also get that,

$P(k+1) = (k+1)[(k+1)^{2} + 5]$

$= (k+1)(k^2 + 2k + 6)$

$= k(k^2 + 5 ) + k(2k+1) + k^2+2k+ 6$

$= 6x + 3k^2 + 3k + 6$ $\because (k^2 + k ) \text{ is even }$

$= 6x + 3\times 2y + 6$, is divisible by 6

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural number n it is true that, P(n)= $n(n^2 + 5)$ is divisible by 6.

Question:11

$n^2 < 2^n$ for all natural numbers $n \geq 5$.

Answer:

P(n)=$n^2 < 2^n$ for $n \geq 5$.....(given)

Let us consider

$P(k) = k^2< 2^k$ to be true,

Thus, $P(k+1) = (k+1)^2$

$= k^2 + 2k + 1$

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

Now, since, $n^2> 2n + 1$ for $n\geq 3$

$k^2 + 2k + 1 < 2k^2$

Thus, $(k+1)^2< 2^{k+1}$

Thus, P(k+1) is true

Hence, by mathematical induction,

For each natural no. n2< 2n it is true that, P(n)=$n^2 < 2^n$ for $n \geq 5$.

Question:12

$2n < (n + 2)!$ for all natural number n.

Answer:

$2n < (n + 2)!$ .....(given)

Now, we will substitute different values for n,

$P(0) \rightarrow 0 < 2!$

$P(1) \rightarrow 2 < 3!$

$P(2) \rightarrow 4 < 4!$

$P(3) \rightarrow 6 < 5!$

Now, let us consider,

$P(k) = 2k < (k+2)!$ To be true,

Thus, $P(k+1) = 2(k+1)< [(k+1) + 2]!$

Now, we know that,$[(k+1) + 2]! = (k+3)! = [(k+3) (k+2)(k+1) ... 3 \times 2\times 1]$

But also, $= 2(k+1) \times (k +3)(k+2) ... 3 \times 1 > 2(k+1)$

Thus,$2(k+1) < [(k+1) + 2]!$

Thus, P(k+1) is true.

Hence, by mathematical induction,

For each natural no. n it is true that, P(n) = $2n < (n + 2)!$.

Question:13

$\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{n}}$ for all natural numbers $n \geq 2$.

Answer:

$
\mathrm{P}(n) \text { is } \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{n}} ; n \geq 2 .
$

$\mathrm{P}(2)$ is $\sqrt{2}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}=1.414<1.7 .0 \ldots$ thus, itistrue $\mathrm{P}(3)$ is $\sqrt{3}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}=1.732<2.284$, thus itis true Now, letusconsider $P(k)=\sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}}$, is true Now, wewilladd $\sqrt{k+1}$ on both sides,

Thus,

$
\sqrt{k}+\sqrt{k+1}-\sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}}+\sqrt{k+1}-\sqrt{k}
$


Thus, $\sqrt{k+1}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots .+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}$
Hence, by mathematical induction,
For each natural no. $n \geq 2, \mathrm{P}(\mathrm{n})$ is $\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}$.

Question:14

$2 + 4 + 6 +...+ 2n = n^2 + n$ for all natural numbers n.

Answer:P(n)=$2 + 4 + 6 +...+ 2n = n^2 + n$

Now, we will substitute different values for n,

$P(0) = 0 = 0^2 + 0$, is true

$P(1) = 2 = 1^2 + 1$, is true

$P(2) = 2+4 = 2^2 + 2$, is true

Now, let us consider,

$P(k) = 2 + 4 + 6 + .... + 2k = k^2 + k$ to be true,

Thus,

P(k+1) is$2 + 4 + 6 + .... + 2k + 2(k+1) = k^2 + k + 2k + 2$

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

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

Thus, P(k+1) is true if P(k) is true

Hence, by mathematical induction,

For each natural number n, it is true that P(n)=$2 + 4 + 6 +...+ 2n = n^2 + n$.

Question:15

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

Answer:

P(n) is $1 + 2 + 2^2 + ... 2^n = 2^{n+1} -1$ ..................given

Now, we’ll substitute different values for n,

$P(0) = 1 = 2^{0+1} -1$ , is true

$P(1) = 1 + 2 = 3 = 2^{1+1} -1$ , is true

$P(2) = 1 + 2+ 2^2 = 7 = 2^{2+1} -1$, is true

Now, let us consider,

$P(k) = 1 + 2 + 2^2 + .... 2^k = 2^{k+1} -1$ to be true

Thus,

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

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

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

Thus, P(k+1) is true if P(k) is true

Hence, by mathematical induction,

For each natural number n it is true that, P(n) is $1 + 2 + 2^2 + ... 2^n = 2^{n+1} -1$.

Question:16

$1 + 5 + 9 + ...+(4n -3) = n (2n -1)$ for all natural number n.

Answer:

P(n)= $1 + 5 + 9 + ...+(4n -3) = n (2n -1)$

Now, we’ll substitute different values for n,

P(1) = 1

= 1(2-1), is true

P(2) = 1+5 = 6

= 2(4-1), is true

Now, let us consider,

$P(k) = 1+5+9+.....+[4k-3] = k(2k-1)$
$P(k+1) = 1+5+9+.....+[4(k+1)-3] = k(2k-1) + 4(k+1)-3$

$= k(2k -1) + 4k + 1\\ (k+1)[2(k+1)-1]$

Thus, P(k+1) is true if P(k) is true

Hence, by mathematical induction,

For each natural no. n it is true that, P(n)= $1 + 5 + 9 + ...+(4n -3) = n (2n -1)$

Question:17

A sequence $a_1, a_2, a_3 ...$ is defined by letting $a_1 = 3 \ and \ a_k = 7a_{k-1}$ for all natural numbers k ≥ 2. Show that $a_n = 3.7^{n-1 }$for all natural numbers.

Answer:

Given

$
a_1=3 \& a_k=7 a_{k-1}
$

Now, we'll substitute different values for $k$

$
\begin{aligned}
& a_2=7 \times a_1 \\
&=7 \times 3=21 \\
&=3.7^{2-1} \\
& a_3=7 \times a_2 \\
&=7^2 \times a_1=49 \times 3 \\
&=147 \\
&=3.7^{3-1} \\
& a_n=7 a_{n-1} \\
&=3.7^{n-1}
\end{aligned}
$

Thus, $a_{n+1}=7 a_{n+1-1}$

$
\begin{aligned}
& =7 a_n \\
& =7 \times 3.7^{n-1} \\
& =3.7^{(m+1)-1}
\end{aligned}
$

Thus, $a_{n+1}$ is true if $a_n$ is true
Hence, by mathematical induction,
For each natural number. It is true that $a_n=3.7^{n-1}$

Question:18

A sequence b0, b1, b2 ... is defined by letting $b_0=5$ and $b_k = 4 + b_{k -1}$ for all natural numbers k. Show that for all natural numbers n, $b_n = 5 + 4n$ using mathematical induction.

Answer:

Given:

The sequence b0, b1, b2,... if defined if we let, $b_0 = 5$& for all natural numbers k. $b_k = 4 +b_{k-1}$Thus,

$b_1 = 4 + b_0\\ = 4 + 5\\ = 9 \\ =5 + 4.1$

$b_2 = 4 + b_1\\ = 4 + 9\\ = 13\\ = 5 + 4.2\\$

Now, let us consider,$b_m = 4 + b_{m-1} = 5 + 4m$ to be true.

Thus,

$b_{m+1} = 4 + b_m = 4 + 5+ 4m \\= 5 + 4(m+1)\\$

Thus, bm+1 is true if bm is true

Hence, by mathematical induction,

For each natural number n it is true that $b_n = 5 + 4n$

Question:19

A sequence $d_1, d_2, d_3 \ldots$. Is defined by letting $\mathrm{d}_1=2$ and $\mathrm{d}_{\mathrm{k}}=\frac{\mathrm{d}_{\mathrm{k}-1}}{\mathrm{k}} \mathrm{d}_{\mathrm{n}}=\frac{2}{\mathrm{n}!}$ numbers, $k \geq 2$. Show that $\mathrm{d}_{\mathrm{n}}=\frac{2}{\mathrm{n}!}$ for all $\mathrm{n} \in \mathrm{N}$..

Answer:

A sequence $\mathrm{d}_1, \mathrm{~d}_2, \mathrm{~d}_3 \ldots \ldots$ Is defined by letting

$
\begin{aligned}
& \mathrm{d}_1=2 \text { and } \mathrm{d}_{\mathrm{k}}=\frac{\mathrm{d}_{\mathrm{k}-1}}{\mathrm{k}} \mathrm{~d}_{\mathrm{n}}=\frac{2}{\mathrm{n}!} \text { numbers, } k \geq 2 \\
& \mathrm{~d}_k=\mathrm{d}_{k-1} / \mathrm{k}
\end{aligned}
$

$\mathrm{d}_2=\mathrm{d}_1 / 2$

$=2 / 2=2 / 2!$

$\mathrm{d}_3=\mathrm{d}_2 / 3$

$=1 / 3=2 / 3!$

Now, let us consider,

$
\mathrm{d}_m=\mathrm{d}_{m-1} / \mathrm{m}=2 / \mathrm{m}!\text { to be true. }
$

Thus,

$
\begin{aligned}
& d_{m+1}=d_{m+1-1} / m+1 \\
& =\mathrm{d}_{\mathrm{m}} / \mathrm{m}+1 \\
& =2 /(\mathrm{m}+1) \cdot \mathrm{m}! \\
& =2 /(\mathrm{m}+1)!
\end{aligned}
$

Thus, $d_{m+1}$ is true if $d_m$ is true
Hence, by mathematical induction,
For each natural no. n it is true that, $\mathrm{d}_{\mathrm{n}}=\frac{2}{\mathrm{n}!}$

Now, we'll substitute different values for $k$.

Question:20

Prove that for all n $\epsilon$ N
${c} \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(n-1) \beta) \\\\ $$=\frac{\cos \left(\alpha+\left(\frac{n-1}{2}\right) \beta ) \sin \left(\frac{n \beta}{2}\right)\right.}{\sin \frac{\beta}{2}} $

Answer:

Given data:

$
\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(n-1) \beta)
$
$=\frac{\cos \left(\alpha+\left(\frac{n-1}{2}\right) \beta\right) \sin \left(\frac{n \beta}{2}\right)}{\sin \frac{\beta}{2}}$

Now, we'll substitute different values for n ,

$
\begin{aligned}
& \mathrm{n}=1 \\
& \cos (\alpha+(1-1) \beta)=\cos \alpha
\end{aligned}
$
$=\frac{\cos \left(\alpha+\left(\frac{1-1}{2}\right) \beta\right) \sin \left(\frac{\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$

Thus, it's true

Now, $n=2$,

$
\cos \alpha+\cos (\alpha+(2-1) \beta)=\cos \alpha+\cos (\alpha+\beta)
$
$=\frac{\cos \left(\alpha+\left(\frac{2-1}{2}\right) \beta\right) \sin \left(\frac{2 \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$

$2 \cos (\alpha+\beta / 2) \cos (\beta / 2)$

Thus, it's true.

$
\text { Now, let } \mathrm{n}=\mathrm{k} \\
{c}
\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(k-1) \beta)$

$=\frac{\cos \left(\alpha+\left(\frac{k-1}{2}\right) \beta\right) \sin \left(\frac{k \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} \text { be true } \\$

$\text { Thus, } n=k+1 \\
\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(k-1) \beta)+\cos (\alpha+(k+1-1) \beta)\\$

$=\frac{\cos \left(\alpha+\left(\frac{k-1}{2}\right) \beta\right) \sin \left(\frac{k \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}+\cos (\alpha+k \beta) \\$
$=\frac{\cos \left(\alpha+\left(\frac{k-1}{2}\right) \beta\right) \sin \left(\frac{k \beta}{2}\right)+\cos (\alpha+k \beta) \sin \left(\frac{\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$
$=\frac{\sin \left(\alpha+\left(\frac{2 k-1}{2}\right) \beta\right) \sin \left(\alpha-\frac{\beta}{2}\right)+\sin (\alpha+(2 k-1) \beta / 2) \sin \left(\alpha+(2 k-1) \frac{\beta}{2}\right)}{2 \sin \left(\frac{\beta}{2}\right)}$

$ \frac{\sin \left(\alpha+\left(\frac{2 k-1}{2}\right) \beta\right) \sin \left(\alpha-\frac{\beta}{2}\right)}{2 \sin \left(\frac{\beta}{2}\right)} \\$
$ =\frac{\cos \left(\alpha+\left(\frac{k}{2}\right) \beta\right) \sin \left(\frac{(k-1) \beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)}$
Thus, $\mathrm{n}=\mathrm{k}+1$ is true
Thus, by mathematical Induction,
For each natural no. n it is true that,

$
\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos (\alpha+(n-1) \beta)
$
$=\frac{\cos \left(\alpha+\left(\frac{n-1}{2}\right) \beta\right) \sin \left(\frac{n \beta}{2}\right)}{\sin \frac{\beta}{2}}$

Question:21

Prove that, $\cos \theta \cos 2 \theta \cos 2^{2} \theta...\cos 2^{n-1} \theta=\frac{\sin 2^{n} \theta}{2^{n} \sin \theta}$, for all n $\epsilon$N.

Answer:

Given:

$
\cos \theta \cos 2 \theta \cos 2^2 \theta \ldots \cos 2^{n-1} \theta=\frac{\sin 2^n \theta}{2^n \sin \theta}
$


Now, we'll substitute different values for n,

n = 1,

$\cos \theta=2 \sin \theta \cos \theta / 2 \sin \theta$

$=\sin 2^1 \theta / 2^1 \sin \theta$

Thus, it's true.
Now, $n=2$

$\cos \theta \cos 2 \theta=\sin 2^2 \theta / 2^2 \sin \theta$

$\begin{aligned} & =2 \sin 2 \theta \cos 2 \theta / 4 \sin \theta \\ & =4 \sin \theta \cos \theta \cos 2 \theta / 4 \sin \theta\end{aligned}$

Thus, it is true.

$
\begin{aligned}
& \quad n=3 \\
& \cos \theta \cos 2 \theta \cos 4 \theta=\sin 2^3 \theta / 2^3 \sin \theta \\
& =2 \sin 4 \theta \cos 4 \theta / 8 \sin \theta \\
& =4 \sin 2 \theta \cos 2 \theta \cos 4 \theta / 8 \sin \theta \\
& =8 \sin \theta \cos \theta \cos 2 \theta \cos 4 \theta / 8 \sin \theta
\end{aligned}
$

n = k,

$\cos \theta \cos 2 \theta \cos 2^2 \theta \ldots \cos 2^{\kappa-1} \theta=\sin 2^\kappa \theta / 2^\kappa \sin \theta$ to be true.
Thus, at $n=k+1$,

$=\cos \theta \cos 2 \theta \cos 2^2 \theta \ldots \cos 2^{k-1} \theta \cos 2^{k+1-1} \theta$

$=\sin 2^k \theta / 2^k \sin \theta \cdot \cos 2^k \theta$
$=2 \sin 2^k \theta \cos 2^k \theta / 2^{k+1} \sin \theta \\ $

$ =\sin 2^{k+1} \theta / 2^{k+1} \sin \theta$

Thus, $n=k+1$ is true
Thus, by mathematical Induction,
For each natural no. n it is true that,

$\cos \theta \cos 2 \theta \cos 2^2 \theta \ldots \cos 2^{n-1} \theta=\frac{\sin 2^n \theta}{2^n \sin \theta}$

Question:22

Prove that, $\sin \theta+\sin 2\theta+\sin 3\theta+...+\sin n\theta\begin{aligned} &=\frac{\frac{\sin \mathrm{n} \theta}{2} \sin \frac{(\mathrm{n}+1)}{2} \theta}{\sin \frac{\theta}{2}} \end{aligned}$ for all $n \in N$

Answer:

Given:

P(n) : $\sin \theta+\sin 2\theta+\sin 3\theta+...+\sin n\theta\begin{aligned} &=\frac{\frac{\sin \mathrm{n} \theta}{2} \sin \frac{(\mathrm{n}+1)}{2} \theta}{\sin \frac{\theta}{2}} \end{aligned}$
Now, we’ll substitute different values for n=1,
$\\\sin \theta=[\sin (1 \theta / 2) \sin (1+1) \theta / 2] / \sin \theta / 2$
Thus, it is true.Now,
let us assume that, for n=k is true
$\sin \theta+\sin 2 \theta+\sin 3 \theta+\ldots \ldots+\sin k \theta=[\sin (k \theta / 2) \sin (k+1) \theta / 2] / \sin \theta / 2$
At n = k+1
$\sin \theta+\sin 2 \theta+\sin 3 \theta+\ldots \ldots+\sin k \theta+\sin (k+1) \theta\\$
$\\= \sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \theta / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+1) \theta / 2+\cos (2 k+1) \theta / 2-\cos (2 k+3) \theta / 2] / 2 \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+3) \Theta / 2] / 2 \sin \theta / 2$ $\\ =[\sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \ominus / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2$
$\\ =\sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \theta / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+1) \theta / 2+\cos (2 k+1) \theta / 2-\cos (2 k+3) \theta / 2] / 2 \sin \theta / 2$ $\\ =[\cos (\theta / 2)-\cos (2 k+3) \Theta / 2] / 2 \sin \theta / 2$ $\\ =[\sin \mathrm{k} \theta / 2 \sin (\mathrm{k}+1) \ominus / 2+\sin \theta / 2 \sin (\mathrm{k}+1) \Theta] / \sin \theta / 2$
Thus, n=k+1 is true
Thus, by mathematical Induction,
For each natural no. n it is true that,
$\sin \theta+\sin 2\theta+\sin 3\theta+...+\sin n\theta\begin{aligned} &=\frac{\frac{\sin \mathrm{n} \theta}{2} \sin \frac{(\mathrm{n}+1)}{2} \theta}{\sin \frac{\theta}{2}} \end{aligned}$

Question:23

Show that $\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7n}{15}$ is a natural number for all n $\epsilon$N.

Answer:

Given data:
P(n) = $\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7n}{15}$
Now, we’ll substitute different values for n,
$\text { At } n=1 $

$\\ P(1)=\frac{1^{5}}{5}+\frac{1^{8}}{3}+\frac{7(1)}{15} \\ $

$=\frac{1}{5}+\frac{1}{3}+\frac{7}{15}$
$\\=(3+5+7 )/ 15\\ =1 \: \: is\: \: true.\\\\\ $

$At \: \:$

$n=2$

$ P(2)=\frac{2^{5}}{5} +\frac{2^{3}}{3}+\frac{7(2)}{15} \\\\$

$ =\frac{32}{5}+\frac{8}{3}+\frac{14}{15} \\\\ =(96+40+14 )/ 15 =10 \: \: is\: \: true$
At n=3
$\\P(3)=\frac{3^{5}}{5} +\frac{3^{3}}{3}+\frac{7(3)}{15} \\\\ $

$=\frac{243}{5}+\frac{27}{3}+\frac{21}{15} \\\\ =(729+135+21 )/ 15 \\\\$

$=59$ is true.
Let us consider,
$P(k)=\frac{k^{5}}{5}+\frac{k^{2}}{3}+\frac{7(k)}{15}$$\: \: to\: \: be\: \: true.$
Thus, at n=k+1
$ P(k+1)=\frac{(k+1)^5}{5}+\frac{(k+1)^3}{3}+\frac{7(k+1)}{15} \\ $

$ =\frac{k^5+5 k^4+10 k^2+10 k^2+5 k+1}{5}+\frac{k^3+3 k^2+3 k+1}{3}+\frac{7(k+1)}{15} \\ $

$=\frac{k^5}{5}+\frac{k^3}{3}+\frac{7(k)}{15}+k^4+2 k^3+3 k^2+2 k+1, \text { is a natural number }$
Thus, n = k + 1 is true
Thus, by mathematical Induction,
For each natural number. It is true that,
P(n) = $\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7n}{15}$

Question:24

Prove that $\frac{1}{n+1}+\frac{1}{n+2}+\ldots .+\frac{1}{2 n}>\frac{13}{24}$ ,for all natural numbers n > 1.

Answer:

Given:
P(n) = $\frac{1}{n+1}+\frac{1}{n+2}+\ldots .+\frac{1}{2 n}>\frac{13}{24}$
Now, we’ll substitute different values for n,
$ \text { At } n=2 \\$

$ P(2)=1 / 2 +1 / 4 \\ =3 / 4>13 / 24 \\$

$ It\: \: is\: \: true.\\ $

$At \: \: $n=3$ \\ $

$P(3)=1 / 2+1 / 4+1 / 6 \\ =11 / 12>13 / 24 \\$.
It is true.
Now, let us consider,

$P(k)=\frac{1}{k+1}+\frac{1}{k+2}+\ldots \ldots+\frac{1}{2 k}>\frac{13}{24}$ To be true.
Now, at $n=k+1$,

$
P(k+1)=\frac{1}{(k+1)+1}+\frac{1}{(k+1)+2}+\ldots . .+\frac{1}{2(k+1)}
$

$\begin{array}{l} =\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k+2} \\\\ =\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k}+\frac{1}{2 k+1}+\frac{1}{2 k+2}-\frac{1}{k+1} \\\\ =\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k}+\frac{2k+2+2k+1-4k-2}{2(2 k+1)(k+1)} \\\\ =\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3} \ldots \ldots+\frac{1}{2 k}+\frac{1}{2(2 k+1)(k+1)}>\frac{13}{24} \end{array}$
Thus, n = k + 1 is true
Thus, by mathematical Induction,
For each natural no. n > 1 it is true that,
P(n) = $\frac{1}{n+1}+\frac{1}{n+2}+\ldots .+\frac{1}{2 n}>\frac{13}{24}$

Question:25

Prove that the number of subsets of a set containing n distinct elements is $2^n$, for all n $\epsilon$ N.

Answer:

To Prove:
$2^n$ is the no. of subsets of a set containing n distinct elements.
Now, there is only one subset since for a null set, there is only one element φ
Now, we’ll substitute different values for n,
At n = 0,
No. of subsets = 1 = $2^0$
If we consider a set that has 1 element, then there will be 2 subsets that have 1 & φ
Thus, at n = 1, it is true
now at n = 2
No. of subsets = 2 = $2^2$
Now, let us consider that there is a set $S_k$ that has k elements.
At n = k,
No. of subsets = $2^k$ is true.
Now, for set $S_{k+1}$, having k+1 elements,
Compared to the set $S_{k}$, $S_{k+1}$ has an extra element and thus will form an extra collection of subsets when combined with the existing $2^k$ subsets & an extra $2^k$ subset will be formed.
Thus, at n = k+1,
No. of subsets = $2^k$ + $2^k$
= 2.$2^k$
= $2^{k+1}$
Thus, by mathematical Induction,
$2^n$ is the no. of subsets of a set containing n distinct elements.

Question:26

If $10^n + 3.4^{n+2} + k$ is divisible by 9 for all n? N, then the least positive integral value of k is
A. 5
B. 3
C. 7
D. 1

Answer:

The answer is option (A)

Given:

P(n) = $10^n + 3.4^{n+2} + k$ is divisible by 9

Now, we’ll substitute different values for n,

At n = 1,

$P(1) = 10^1 + 3.4^3+ k$

$= 202 + k$

Now, we know that the sum of the digits of this no should be 9 or multiples of 9, for it to be divisible by 9.

Thus $2 + 2 + k = 9$

Thus,$k = 5.$

Question:27

For all n? N, $3.5^{2n + 1} + 2^{3n + 1}$ is divisible by
A. 19
B. 17
C. 23
D. 25

Answer:

The answer is option (B)
Given:
P(n) = $3.5^{2n + 1} + 2^{3n + 1}$
Now, we’ll substitute different values for n,
At n = 1,
P(1) = $3.5^{2+1} + 2^{3+1}$
$\\ = 375 + 16\\ = 391 \\=17\times 23$
At n = 2,
$P(2) = 3.5^{4+1} + 2^{6+1}$
$\\= 9375 + 256 \\ = 9503 \\ = 17 \times 559$
Therefore, it is divisible by 17.

Question:28

If $x^n-1$ is divisible by$x - k$ , then the least positive integral value of k is
A. 1
B. 2
C. 3
D. 4

Answer:

The answer is option (A)
Given:
P(n) = $x^n-1$ is divisible by$x - k$
Now, we’ll substitute different values for n,
At n = 1,
$P(1) = x -1$
At n = 2,
$P(2) = x^2 -1$
$= (x -1) (x + 1)$
At n = 3,
$P(3) = x^3 -1$
$= (x -1) (x^2 + xy + 1)$
At n = 4,
$\\P(4) = x^4 -1\\ = (x^2 -1) (x^2 + 1) \\ = (x -1) (x + 1) (x^2 + 1)$

Therefore, ‘1’ is the least possible integral value of k.

Question:29

Fill in the blanks in the following:

If $P(n) : 2n < n!, n \in N$, then P(n) is true for all $n\geq$ __________.

Answer:

$n\geq4$
Given:
$P(n) : 2n < n!, n \in N$
Now, we’ll substitute different values for n,
At n = 1,
$P(1) = 2 \times 1 < 1!$,
$= 2 < 1$, viz. not true.
At n = 2,
$P(2) = 2 \times 2 < 2!$
$= 4 < 2$, viz. not true.
At n = 3,
$P(3) = 2\times 3 < 3!$
$= 6 < 6$, viz. not true
At n = 4,
$P(4) = 2 \times 4 < 4!$
$= 8 < 24$, viz. not true
At n = 5,
$P(5) = 2 \times 5 < 5!$
$= 10 < 120$, viz. true
Therefore, $n\geq4$

Subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 4

  • Introduction
  • Motivation
  • The Principle of Mathematical Induction

Importance of solving NCERT Exemplar Class 11 Maths Questions

These solutions will ultimately help students develop skills regarding inductive and deductive reasoning used in almost every other field, ranging from science to many other branches.

  • The Solutions for Class 11 NCERT Exemplar Mathematics are simple to understand and provide a wide range of questions to practice.
  • After solving these questions, students will be able to grasp the underlying concepts behind each solution, which enhances their understanding and resolves their uncertainties about tackling similar problems.
  • Since these solutions are available online, you can practice these questions at your convenience.

NCERT solutions of class 11 - Subject-wise

Here are the subject-wise links for the NCERT solutions of class 11:

NCERT Notes of class 11 - Subject Wise

Given below are the subject-wise NCERT Notes of class 11 :

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 11:

NCERT Exemplar Class 11 Solutions

Given below are the subject-wise exemplar solutions of class 11 NCERT:

Frequently Asked Questions (FAQs)

Q: What is the Principle of Mathematical Induction in Class 11 NCERT Exemplar?
A:

To prove that a given statement is true for all natural numbers, the Principle of Mathematical Induction is applied. It is based on two essential steps:

  • basic event (In the base event, we prove that the statement is true for the most minimal value, typically n = 1)
  • inductive step. (In the inductive step, we assume that the statement contains a couple of arbitrary values.)
Q: How to solve NCERT Exemplar problems for Chapter 4 Class 11 Maths?
A:

To answer questions on chapter 4 (Mathematical Induction ) of the NCERT Exemplar, first, carefully read the question to find out what's on the other side. Get down to business by proving the basic case, where you check that the statement is correct for a small amount.

n, often 

n=1. Next, assume that the statement is true for some 

n=k, that's the inductive theory. In the Inductive move, use the presumption to show that the assertion includes.

n=k+2. If you succeed in proving it, you have revealed that this observation applies to all inherent numbers. To solve these problems, you need practice, as you need to carefully work by taking all the steps to ensure that the initiation method is properly followed.

Q: What are the key concepts covered in Chapter 4 of NCERT Exemplar Class 11 Maths?
A:

This chapter includes solving a result using the principle of mathematical induction. It uses the basic situation to derive the result and then applies it for higher values.

Q: Can you explain the step-by-step process of mathematical induction?
A:

The process of Mathematical Induction follows a structured, step-by-step approach. First, you begin by proving the Base Case, where you demonstrate that the statement is true for the smallest value, typically n=1. Next, in the Inductive Hypothesis, you assume that the statement is true for some arbitrary value n=k. This assumption is then used to prove the Inductive Step, where you show that the statement holds for n=k+1. Once you have successfully proven both the base case and the inductive step, you can conclude, by the principle of induction, that the statement is true for all natural numbers. The step-by-step nature of this process ensures that the proof is rigorous and logical.

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