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 $n \geq 4$ but $P(1), P(2) \: \: and \: \: P(3)$ are $\text{Not True}$. Justify your answer.

Answer:

Given:
$P(n)$ which it true for all $n \geq 4$ but $P(1), P(2) \: \: and \: \: P(3)$ are $\text{Not True}$.

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

Examples of the given statements are-

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

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

$P(2) = 2^2<2!$ i.e., $4<2 \rightarrow$ Not True
$P(3) = 2^3<3!$ i.e., $8<6 \rightarrow$ Not True
$P(4) = 2^4<4!$ 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,

$1+ 2 + 3 + 4 +... + n = \frac{n (n+1)}2$
P(0) $=\frac{0(0+1)}2 = 0 $

$\rightarrow \text{true for} \ P (1) $

$\rightarrow1 = \frac{1(2+2)}2 = 1 $

$\rightarrow \text{true for}\ P (2)$

$ \rightarrow 1+2 = \frac{2(2+1)}2 $

$\rightarrow \text{true for}\ P (k) $

$\rightarrow 1+2+3+... +k = \frac{k (k+1)}2$
For $P (k+1) $

$\rightarrow 1+2+3+... +k+1 = \frac{k (k+1)}2 + k + 1 = \frac{(k+1) (k+2)}2$
$ \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 number 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 number 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 number 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 number 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 number $n \geq 2, \mathrm{P}(\mathrm{n})$ 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 number 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 number 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 number 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 number 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 number 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 number 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 number 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,
number 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
number of subsets = 2 = $2^2$
Now, let us consider that there is a set $S_k$ that has k elements.
At n = k,
number 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,
number of subsets = $2^k$ + $2^k$
= 2.$2^k$
= $2^{k+1}$
Thus, by mathematical Induction,
$2^n$ is the number 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, $n\geq4$

Question 30

True or False
Let P(n) be a statement and let $P(k) \Rightarrow P(k + 1)$, for some natural number k, then P(n) is true of all n ? N.

Answer:

True.

Explanation: It is the Principle of Mathematical Induction.