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

Question:1

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

Answer:

Given:
$P(n)$ which it true for all $n\ge 4$ but $P(1),P(2)andP(3)$ 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., $1<1\to NotTrue$

$P(1)=2^1<1!$ i.e., $2<1\to NotTrue$

$P(2)=2^2<2!$ i.e., $4<2\to$ NotTrue
$P(3)=2^3<3!$ i.e., $8<6\to$ NotTrue
$P(4)=2^4<4!$ i.e., $16<24\to$ True
$\mathrm{\&}P(5)=2^5<5!$ i.e., $32<60\to$ 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) $=0(0+1)/2=0$

$$\begin{gathered} \to true \\ P(1) \end{gathered}$$

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

$$\begin{gathered} \to true \\ P(2) \end{gathered}$$

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

$$\begin{gathered} \to true \\ P(k) \end{gathered}$$

$$\begin{gathered} \to 1+2+3+...+k=k(k+1)/2 \\ P(k+1) \end{gathered}$$

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

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,

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

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

$P(k+1)=2^{3(k+1)}-1=2^3(3x+1)-1=56x+7,\text{ is divisible by }7$
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$x\ne y$.

Answer:

P(n)= $x^n-y^n$ is divisible by x -y, where x integers with$x\ne y$. ........(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

$\begin{array}{rl}& P(k+1)=x^{k+1}-y^{k+1}\\ & =x^k(x-y)+y(x^k-y^k)\\ & =x^k(x-y)+ya(x-y),\text{ is divisible by }\mathrm{x}-\mathrm{y}\end{array}$
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 $\mathrm{x}\ne \mathrm{y}$.

Question:9

$n^3-n$ is divisible by 6, for each natural number n. $n\ge 2$

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 $n\ge 5$.

Answer:

P(n)=$n^2<2^n$ for $n\ge 5$.....(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 $n\ge 3$

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

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 $n\ge 5$.

Question:12

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

Answer:

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

Now, we will substitute different values for n,

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

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

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

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

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

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

Answer:

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

$\mathrm{P}(2)$ is $\sqrt{2}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}=1.414<1.7.0\dots$ 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 $P(k)=\sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots +\frac{1}{\sqrt{k}}$, is true Now, wewilladd $\sqrt{k+1}$ on both sides,

Thus,

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

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

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,

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

$$\begin{gathered} =k(2k-1)+4k+1 \\ (k+1)[2(k+1)-1] \end{gathered}$$

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 $a_1=3and{a}_k=7a_{k-1}$ for all natural numbers k ≥ 2. Show that $a_n={3.7}^{n-1}$for all natural numbers.

Answer:

Given

$a_1=3\mathrm{\&}{a}_k=7a_{k-1}$

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

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

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

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

$$\begin{gathered} b_1=4+b_0 \\ =4+5 \\ =9 \\ =5+4.1 \end{gathered}$$

$$\begin{gathered} b_2=4+b_1 \\ =4+9 \\ =13 \\ =5+4.2 \end{gathered}$$

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

Thus,

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

Thus, b_m+1 is true if b_m is true

Hence, by mathematical induction,

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

Question:19

A sequence $d_1,d_2,d_3\dots$. 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, $k\ge 2$. Show that ${\mathrm{d}}_{\mathrm{n}}=\frac{2}{\mathrm{n}!}$ for all $\mathrm{n}\in \mathrm{N}$..

Answer:

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

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

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

$=2/2=2/2!$

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

$=1/3=2/3!$

Now, let us consider,

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

Thus,

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

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 $ϵ$ N
$c\cos \alpha +\cos (\alpha +\beta )+\cos (\alpha +2\beta )+\dots +\cos (\alpha +(n-1)\beta )$$=\frac{\cos (\alpha +\left(\frac{n-1}{2}\right)\beta )\sin \left(\frac{n\beta }{2}\right)}{\sin \frac{\beta }{2}}$

Answer:

Given data:

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

Now, we'll substitute different values for n ,

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

Thus, it's true

Now, $n=2$,

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

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

Thus, it's true.

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

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

$$\begin{gathered} \text{ Thus, }n=k+1 \\ \cos \alpha +\cos (\alpha +\beta )+\cos (\alpha +2\beta )+\dots +\cos (\alpha +(k-1)\beta )+\cos (\alpha +(k+1-1)\beta ) \end{gathered}$$

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

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

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

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 $ϵ$N.

Answer:

Given:

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

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, $n=2$

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

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

Thus, it is true.

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

n = k,

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

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

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

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

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

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

Question:22

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

Answer:

Given:

P(n) : $\sin \theta +\sin 2\theta +\sin 3\theta +...+\sin n\theta \begin{array}{rl}& =\frac{\frac{\sin \mathrm{n}\theta }{2}\sin \frac{(\mathrm{n}+1)}{2}\theta }{\sin \frac{\theta }{2}}\end{array}$
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
$\sin \theta +\sin 2\theta +\sin 3\theta +\dots \dots +\sin k\theta =[\sin (k\theta /2)\sin (k+1)\theta /2]/\sin \theta /2$
At n = k+1
$\sin \theta +\sin 2\theta +\sin 3\theta +\dots \dots +\sin k\theta +\sin (k+1)\theta$
$=\sin \mathrm{k}\theta /2\sin (\mathrm{k}+1)\theta /2+\sin \theta /2\sin (\mathrm{k}+1)\mathrm{\Theta }]/\sin \theta /2$ $=[\cos (\theta /2)-\cos (2k+1)\theta /2+\cos (2k+1)\theta /2-\cos (2k+3)\theta /2]/2\sin \theta /2$ $=[\cos (\theta /2)-\cos (2k+3)\mathrm{\Theta }/2]/2\sin \theta /2$ $=[\sin \mathrm{k}\theta /2\sin (\mathrm{k}+1)\ominus /2+\sin \theta /2\sin (\mathrm{k}+1)\mathrm{\Theta }]/\sin \theta /2$
$=\sin \mathrm{k}\theta /2\sin (\mathrm{k}+1)\theta /2+\sin \theta /2\sin (\mathrm{k}+1)\mathrm{\Theta }]/\sin \theta /2$ $=[\cos (\theta /2)-\cos (2k+1)\theta /2+\cos (2k+1)\theta /2-\cos (2k+3)\theta /2]/2\sin \theta /2$ $=[\cos (\theta /2)-\cos (2k+3)\mathrm{\Theta }/2]/2\sin \theta /2$ $=[\sin \mathrm{k}\theta /2\sin (\mathrm{k}+1)\ominus /2+\sin \theta /2\sin (\mathrm{k}+1)\mathrm{\Theta }]/\sin \theta /2$
Thus, n=k+1 is true
Thus, by mathematical Induction,
For each natural no. n it is true that,
$\sin \theta +\sin 2\theta +\sin 3\theta +...+\sin n\theta \begin{array}{rl}& =\frac{\frac{\sin \mathrm{n}\theta }{2}\sin \frac{(\mathrm{n}+1)}{2}\theta }{\sin \frac{\theta }{2}}\end{array}$