Solution

Let the statement $P(n): 3 n<n$ !

For $n=1,3 \times 1<1$ ! $\quad$ [false]

For $n=2,3 \times 2<2 ! \Rightarrow 6<2$ $\quad$ [false]

For $n=3,3 \times 3<3 ! \Rightarrow 9<6$ $\quad$ [false]

For $n=4,3 \times 4<4 ! \quad \Rightarrow \quad 12<24$ $\quad$ [true]

For $n=5,3 \times 5<5 ! \Rightarrow 15<5 \times 4 \times 3 \times 2 \times 1 \Rightarrow 15<120$ $\quad$ [true]

Solution

Consider the statement

$ \begin{aligned} & P(n): 1^{2}+2^{2}+3^{2}+\ldots+n^{2} & =\frac{n(n+1)(2 n+1)}{6} \\ \text { For } n=1, & & =\frac{1(1+1)(2 \times 1+1)}{6} \\ \Rightarrow & & 1 =\frac{2(3)}{6} \\ \Rightarrow & & 1=1 \\ \text { For } n=2, & & 1+2^{2} =\frac{2(2+1)(4+1)}{6} \\ \Rightarrow & & 5=\frac{30}{6} \Rightarrow 5=5 \\ \text { For } n=3, & & 1+2^{2}+3^{2} =\frac{3(3+1)(7)}{6}\\ \Rightarrow & & 1+4+9 =\frac{3(4)(7)}{6}\\ \Rightarrow & & 14 = 14 \end{aligned} $

Hence, the given statement is true for all $n$.

Prove each of the statements in the following questions from by the Principle of Mathematical Induction.

Solution

Let $P(n): 4^{n}-1$ is divisible by 3 for each natural number $n$.

Step I Now, we observe that $P(1)$ is true.

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

It is clear that 3 is divisible by 3.

Hence, $P(1)$ is true.

Step II Assume that, $P(n)$ is true for $n=k$

$P(k): 4^{k}-1$ is divisible by 3

$ \qquad \quad x 4^{k}-1=3 q $

Step III Now, to prove that $P(k+1)$ is true.

$ \begin{aligned} P(k+1) & : 4^{k+1}-1 \\ & =4^{k} \cdot 4-1 \\ & =4^{k} \cdot 3+4^{k}-1 \\ & =3 \cdot 4^{k}+3 q \\ & =3(4^{k}+q) \end{aligned} $

Thus, $P(k+1)$ is true whenever $P(k)$ is true.

Hence, by the principle of mathematical induction $P(n)$ is true for all natural number $n$.

Solution

Let $P(n): 2^{3 n}-1$ is divisible by 7

Step I We observe that $P(1)$ is true.

$ P(1): 2^{3 \times 1}-1=2^{3}-1=8-1=7 $

It is clear that $P(1)$ is true.

Step II Now, assume that $P(n)$ is true for $n=k$, $P(k): 2^{3 k}-1$ is divisible by 7 .

$ \Rightarrow \quad 2^{3 k}-1=7 q $

Step III Now, to prove $P(k+1)$ is true.

$ \begin{aligned} P(k+1) & : 2^{3(k+1)}-1 \\ & =2^{3 k} \cdot 2^{3}-1 \\ & =2^{3 k}(7+1)-1 \\ & =7 \cdot 2^{3 k}+2^{3 k}-1 \\ & =7 \cdot 2^{3 k}+7 q \\ & =7(2^{3 k}+q) \end{aligned} $

Hence, $P(k+1)$ : is true whenever $P(k)$ is true.

So, by the principle of mathematical induction $P(n)$ is true for all natural number $n$.

Solution

Let $P(n): n^{3}-7 n+3$ is divisible by 3 , for all natural number $n$.

Step I We observe that $P(1)$ is true.

Hence, $P(1)$ is true.

$ \begin{aligned} P(1) & =(1)^{3}-7(1)+3 \\ & =1-7+3 \\ & =-3, \text { which is divisible by } 3 \end{aligned} $

Step II Now, assume that $P(n)$ is true for $n=k$.

$ \therefore \quad P(k)=k^{3}-7 k+3=3 q $

Step III To prove $P(k+1)$ is true

$ \begin{aligned} P(k+1) & :(k+1)^{3}-7(k+1)+3 \\ & =k^{3}+1+3 k(k+1)-7 k-7+3 \\ & =k^{3}-7 k+3+3 k(k+1)-6 \\ & =3 q+3[k(k+1)-2] \quad \text {[from step II]} \end{aligned} $

Hence, $P(k+1)$ is true whenever $P(k)$ is true.

So, by the principle of mathematical induction $P(n)$ : is true for all natural number $n$.

Solution

Let $P(n): 3^{2 n}-1$ is divisible by 8 , for all natural numbers.

Step I We observe that $P(1)$ is true.

$ \begin{aligned} P(1): 3^{2(1)}-1 & =3^{2}-1 \\ & =9-1=8, \text { which is divisible by } 8 . \end{aligned} $

Step II Now, assume that $P(n)$ is true for $n=k$.

$ P(k): 3^{2 k}-1=8 q $

Step III Now, to prove $P(k+1)$ is true.

$ \begin{aligned} P(k+1) & : 3^{2(k+1)}-1 \\ & =3^{2 k} \cdot 3^{2}-1 \\ & =3^{2 k} \cdot(8+1)-1 \\ & =8 \cdot 3^{2 k}+3^{2 k}-1 \\ & =8 \cdot 3^{2 k}+8 q \\ & =8(3^{2 k}+q) \end{aligned} $ $\quad$ [from step II]

Hence, $P(k+1)$ is true whenever $P(k)$ is true.

So, by the principle of mathematical induction $P(n)$ is true for all natural numbers $n$.

Solution

Consider the given statement is

$P(n): 7^{n}-2^{n}$ is divisible by 5 , for any natural number $n$.

Step I We observe that $P(1)$ is true.

$ P(1)=7^{1}-2^{1}=5 \text {, which is divisible by } 5 \text {. } $

Step II Now, assume that $P(n)$ is true for $n=k$.

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

Step III Now, to prove $P(k+1)$ is true,

$ \begin{gathered} P(k+1): 7^{k+1}-2^{k+1} \\ =7^{k} \cdot 7-2^{k} \cdot 2 \end{gathered} $

$ \begin{aligned} & =7^{k} \cdot(5+2)-2^{k} \cdot 2 \\ & =7^{k} \cdot 5+2 \cdot 7^{k}-2^{k} \cdot 2 \\ & =5 \cdot 7^{k}+2(7^{k}-2^{k}) \\ & =5 \cdot 7^{k}+2(5 q) \end{aligned} $

$ =5(7^{k}+2 q) \text {, which is divisible by } 5 \text {. [from step II] } $

So, $P(k+1)$ is true whenever $P(k)$ is true.

Hence, by the principle of mathematical induction $P(n)$ is true for any natural number $n$.

Solution

Let $P(n): x^{n}-y^{n}$ is divisible by $x-y$, where $x$ and $y$ are any integers with $x \neq y$.

Step I We observe that $P(1)$ is true.

$ P(1): x^{1}-y^{1}=x-y $

Step II Now, assume that $P(n)$ is true for $n=k$.

$\quad P(k): x^{k}-y^{k}$ is divisible by (x-y) .

$\therefore \quad x^{k}-y^{k}$ = q(x-y)

Step III Now, to prove $P(k+1)$ is true.

$ \begin{aligned} P(k+1) & : x^{k+1}-y^{k+1} \\ & =x^{k} \cdot x-y^{k} \cdot y \\ & =x^{k} \cdot x-x^{k} \cdot y+x^{k} \cdot y-y^{k} \cdot y \\ & =x^{k}(x-y)+y(x^{k}-y^{k}) \\ & =x^{k}(x-y)+y q(x-y) \\ & =(x-y)[x^{k}+y q], \text { which is divisible by }(x-y) . \quad \text { [from step II] } \end{aligned} $

Hence, $P(k+1)$ is true whenever $P(k)$ is true. So, by the principle of mathematical induction $P(n)$ is true for any natural number $n$.

Solution

Let $P(n): n^{3}-n$ is divisible by 6 , for each natural number $n \geq 2$.

Step I We observe that $P(2)$ is true. $P(2):(2)^{3}-2$

$\Rightarrow \qquad 8-2=6 \text {, which is divisible by } 6 \text {. } $

Step II Now, assume that $P(n)$ is true for $n=k$.

$ P(k): k^{3}-k \text { is divisible by } 6 . $

$ \therefore \quad k^{3}-k=6 q $

Step III To prove $P(k+1)$ is true

$ \begin{aligned} P(k & +1):(k+1)^{3}-(k+1) \\ & =k^{3}+1+3 k(k+1)-(k+1) \\ & =k^{3}+1+3 k^{2}+3 k-k-1 \\ & =k^{3}-k+3 k^{2}+3 k \\ & =6 q+3 k(k+1) \end{aligned} $ $\quad$ [from step II]

We know that, $3 k(k+1)$ is divisible by 6 for each natural number $n=k$.

So, $P(k+1)$ is true. Hence, by the principle of mathematical induction $P(n)$ is true.

Solution

Let $P(n): n(n^{2}+5)$ is divisible by 6 , for each natural number $n$.

Step I We observe that $P(1)$ is true.

$P(1): 1(1^{2}+5)=6$, which is divisible by 6 .

Step II Now, assume that $P(n)$ is true for $n=k$.

$P(k): k(k^{2}+5)$ is divisible by 6 .

$\therefore \quad k(k^{2}+5)=6 q$

Step III Now, to prove $P(k+1)$ is true, we have

$ \begin{aligned} P(k+1): & (k+1)[(k+1)^{2}+5] \\ & =(k+1)[k^{2}+2 k+1+5] \\ & =(k+1)[k^{2}+2 k+6] \\ & =k^{3}+2 k^{2}+6 k+k^{2}+2 k+6 \\ & =k^{3}+3 k^{2}+8 k+6 \\ & =k^{3}+5 k+3 k^{2}+3 k+6 \\ & =k(k^{2}+5)+3(k^{2}+k+2) \\ & =(6 q)+3(k^{2}+k+2) \end{aligned} $

We know that, $k^{2}+k+2$ is divisible by 2 , where, $k$ is even or odd.

Since, $P(k+1): 6 q+3(k^{2}+k+2)$ is divisible by 6 . So, $P(k+1)$ is true whenever $P(k)$ is true.

Hence, by the principle of mathematical induction $P(n)$ is true.

Solution

Consider the given statement

$P(n): n^{2}<2^{n}$ for all natural numbers $n \geq 5$.

Step I We observe that $P(5)$ is true

Hence, $P(5)$ is true.

$ \begin{aligned} P(5) & : 5^{2}<2^{5} \\ & =25<32 \end{aligned} $

Step II Now, assume that $P(n)$ is true for $n=k$.

$ P(k)=k^{2}<2^{k} \text { is true. } $

Step III Now, to prove $P(k+1)$ is true, we have to show that

$ P(k+1):(k+1)^{2}<2^{k+1} $

Now,

$ \begin{aligned} k^{2}<2^{k} & =k^{2}+2 k+1<2^{k}+2 k+1 \\ & =(k+1)^{2}<2^{k}+2 k+1 \\ & =2^{k}+2 k+1<2^{k}+2^{k} \\ & =2^{k}+2 k+1<2 \cdot 2^{k} \\ & =2^{k}+2 k+1<2^{k+1} \end{aligned} $

$ \text { Now, }(2 k+1)<2^{k} \quad=2^{k}+2 k+1<2^{k}+2^{k} $

From Eqs. (i) and (ii), we get $(k+1)^{2}<2^{k+1}$

So, $P(k+1)$ is true, whenever $P(k)$ is true. Hence, by the principle of mathematical induction $P(n)$ is true for all natural numbers $n \geq 5$.

Solution

Consider the statement

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

Step I We observe that, $P(1)$ is true. $P(1): 2(1)<(1+2)$ !

$\Rightarrow \qquad 2<3 ! \Rightarrow 2<3 \times 2 \times 1 \Rightarrow 2<6$

Hence, $P(1)$ is true.

Step II Now, assume that $P(n)$ is true for $n=k$,

$P(k): 2 k<(k+2)$ !is true.

Step III To prove $P(k+1)$ is true, we have to show that

$ \qquad \qquad \qquad P(k+1): 2(k+1)<(k+1+2) ! $

Now $\qquad \qquad$ 2 k & <(k+2) !

$\qquad \qquad \qquad$ 2 k <(k+2) !

$\qquad \qquad \qquad$ 2 k+2 <(k+2) !+2

$\qquad \qquad \qquad$ 2(k+1) <(k+2) !+2

$\qquad \qquad \qquad$ (k+2) !+2 <(k+3) !

From Eqs. (i) and (ii),

$ 2(k+1)<(k+1+2) ! $

So, $P(k+1)$ is true, whenever $P(k)$ is true.

Hence, by principle of mathematical induction $P(n)$ is true.

Solution

Consider the statement

$P(n): \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}$, for all natural numbers $n \geq 2$.

Step I We observe that $P(2)$ is true.

$ P(2): \sqrt{2}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}} \text {, which is true. } $

Step II Now, assume that $P(n)$ is true for $n=k$.

$ P(k): \sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k}} \text { is true. } $

Step III To prove $P(k+1)$ is true, we have to show that

Given that,

$ P(k+1): \sqrt{k+1}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{k+1}} \text { is true. } $

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

$\Rightarrow \quad \frac{(\sqrt{k})(\sqrt{k+1})+1}{\sqrt{k+1}}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}$

If

$ \sqrt{k+1}<\frac{\sqrt{k} \sqrt{k+1}+1}{\sqrt{k+1}} $

$ \Rightarrow \quad k+1<\sqrt{k} \sqrt{k+1}+1 $

$ \Rightarrow \quad k<\sqrt{k(k+1)} \Rightarrow \sqrt{k}<\sqrt{k}+1 $

From Eqs. (i) and (ii),

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

So, $P(k+1)$ is true, whenever $P(k)$ is true. Hence, $P(n)$ is true.

Solution

Let $P(n): 2+4+6+\ldots+2 n=n^{2}+n$

For all natural numbers $n$.

Step I We observe that $P(1)$ is true.

$ \begin{aligned} P(1): 2 & =1^{2}+1 \\ 2 & =2, \text { which is true. } \end{aligned} $

Step II Now, assume that $P(n)$ is true for $n=k$.

$ \therefore \quad P(k): 2+4+6+\ldots+2 k=k^{2}+k $

Step III To prove that $P(k+1)$ is true.

$ \begin{aligned} P(k+1) & : 2+4+6+8+\ldots+2 k+2(k+1) \\ & =k^{2}+k+2(k+1) \\ & =k^{2}+k+2 k+2 \\ & =k^{2}+2 k+1+k+1 \\ & =(k+1)^{2}+k+1 \end{aligned} $

So, $P(k+1)$ is true, whenever $P(k)$ is true.

Hence, $P(n)$ is true.

Solution

Consider the given statement

$ P(n): 1+2+2^{2}+\ldots+2^{n}=2^{n+1}-1 \text {, for all natural numbers } n $

Step I We observe that $P(0)$ is true.

$ \begin{aligned} P(1): 1 & =2^{0+1}-1 \\ 1 & =2^{1}-1 \\ 1 & =2-1 \\ 1 & =1, \text { which is true. } \end{aligned} $

Step II Now, assume that $P(n)$ is true for $n=k$.

So, $P(k): 1+2+2^{2}+\ldots+2^{k}=2^{k+1}-1$ is true.

Step III Now, to prove $P(k+1)$ is true.

$ \begin{aligned} P(k+1): & 1+2+2^{2}+\ldots+2^{k}+2^{k+1} \\ & =2^{k+1}-1+2^{k+1} \\ & =2 \cdot 2^{k+1}-1 \\ & =2^{k+2}-1 \\ & =2^{(k+1)+1}-1 \end{aligned} $

So, $P(k+1)$ is true, whenever $P(k)$ is true.

Hence, $P(n)$ is true.

Solution

Let $P(n): 1+5+9+\ldots+(4 n-3)=n(2 n-1)$, for all natural numbers $n$. Step I We observe that $P(1)$ is true.

$ P(1): 1=1(2 \times 1-1), 1=2-1 \text { and } 1=1 \text {, which is true. } $

Step II Now, assume that $P(n)$ is true for $n=k$.

So, $P(k): 1+5+9+\ldots+(4 k-3)=k(2 k-1)$ is true.

Step III Now, to prove $P(k+1)$ is true.

$ \begin{aligned} P(k+1) & : 1+5+9+\ldots+(4 k-3)+4(k+1)-3 \\ & =k(2 k-1)+4(k+1)-3 \\ & =2 k^{2}-k+4 k+4-3 \\ & =2 k^{2}+3 k+1 \\ & =2 k^{2}+2 k+k+1 \\ & =2 k(k+1)+1(k+1) \\ & =(k+1)(2 k+1) \\ & =(k+1)[2 k+1+1-1] \\ & =(k+1)[2(k+1)-1] \end{aligned} $

So, $P(k+1)$ is true, whenever $p(k)$ is true, hence $P(n)$ is true.

Long Answer Type Questions

Use the Principle of Mathematical Induction in the following questions.

Solution

A sequence $a_1, a_2, a_3, \ldots$ is defined by letting $a_1=3$ and $a_{k}=7 a_{k-1}$, for all natural numbers $k \geq 2$.

Let $\quad P(n): a_{n}=3 \cdot 7^{n-1}$ for all natural numbers.

Step I We observe $P(2)$ is true.

Step II Now, assume that $P(n)$ is true for $n=k$.

$ P(k): a_{k}=3 \cdot 7^{k-1} $

Step III Now, to prove $P(k+1)$ is true, we have to show that

$ \begin{aligned} P(k+1): a_{k+1} & =3 \cdot 7^{k+1-1} \\ a_{k+1} & =7 \cdot a_{k+1-1}=7 \cdot a_{k} \\ & =7 \cdot 3 \cdot 7^{k-1}=3 \cdot 7^{k-1+1} \end{aligned} $

So, $P(k+1)$ is true, whenever $p(k)$ is true. Hence, $P(n)$ is true.

Solution

Consider the given statement,

$P(n): b_{n}=5+4 n$, for all natural numbers given that $b_0=5$ and $b_{k}=4+b_{k-1}$

Step I $P(1)$ is true.

$ P(1): b_1=5+4 \times 1=9 $

As $\quad b_0=5, b_1=4+b_0=4+5=9 $

Hence, $P(1)$ is true.

Step II Now, assume that $P(n)$ is true for $n=k$.

$ P(k): b_{k}=5+4 k $

Step III Now, to prove $P(k+1)$ is true, we have to show that

$ \begin{aligned} & \therefore \quad P(k+1): b_{k+1}=5+4(k+1) \\ & b_{k+1}=4+b_{k+1-1} \\ & \qquad =4+b_{k} \\ & \qquad =4+5+4 k=5+4(k+1) \end{aligned} $

So, by the mathematical induction $P(k+1)$ is true whenever $P(k)$ is true, hence $P(n)$ is true.

Solution

Let $P(n): d_{n}=\frac{2}{n !}, \forall n \in N$, to prove $P(2)$ is true.

Step I $\quad P(2): d_2=\frac{2}{2 !}=\frac{2}{2 \times 1}=1 $

As, given $\quad d_1=2 $

$\Rightarrow$ $\quad d_{k}=\frac{d_{k-1}}{k} $

$\Rightarrow \quad d_2=\frac{d_1}{2}=\frac{2}{2}=1$

Hence, $P(2)$ is true.

Step II Now, assume that $P(k)$ is true.

$ P(k): d_{k}=\frac{2}{k !} $

Step III Now, to prove that $P(k+1)$ is true, we have to show that $P(k+1): d_{k+1}=\frac{2}{(k+1) !}$

$ \begin{aligned} d_{k+1} & =\frac{d_{k+1-1}}{k}=\frac{d_{k}}{k} \\ & =\frac{2}{k ! k}=\frac{2}{(k+1) !} \end{aligned} $

So, $P(k+1)$ is true. Hence, $P(n)$ is true.

Solution

Let $P(n): \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(n-1) \beta]$

Step I We observe that $P(1)$

$ =\frac{\cos \alpha+\frac{n-1}{2} \beta \sin \frac{n \beta}{2}}{\sin \frac{\beta}{2}} $

$ P(1): \cos \alpha=\frac{\cos \alpha+\frac{1-1}{2} \quad \beta \sin \frac{\beta}{2}}{\sin \frac{\beta}{2}}=\frac{\cos (\alpha+0) \sin \frac{\beta}{2}}{\sin \frac{\beta}{2}} $

Hence, $P(1)$ is true.

Step II Now, assume that $P(n)$ is true for $n=k$.

$ \begin{aligned} & P(k): \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(k-1) \beta] \\ &= \frac{\cos \alpha+\frac{k-1}{2} \beta \sin \frac{k \beta}{2}}{\sin \frac{\beta}{2}} \end{aligned} $

Step III Now, to prove $P(k+1)$ is true, we have to show that

$P(k+1): \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(k-1) \beta]$

$ +\cos [\alpha+(k+1-1) \beta]=\frac{\cos \alpha+\frac{k \beta}{2} \sin (k+1) \frac{\beta}{2}}{\sin \frac{\beta}{2}} $

LHS $=\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(k-1) \beta]+\cos (\alpha+k \beta)$

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

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

$=\frac{\sin \alpha+\frac{k \beta}{2}-\frac{\beta}{2}+\frac{k \beta}{2}-\sin \alpha+\frac{k \beta}{2}-\frac{\beta}{2}-\frac{k \beta}{2}+\sin \alpha+k \beta+\frac{\beta}{2}-\sin \alpha+k \beta-\frac{\beta}{2}}{2 \sin \frac{\beta}{2}}$

$=\frac{\sin \alpha+k \beta+\frac{\beta}{2}-\sin \alpha-\frac{\beta}{2}}{2 \sin \frac{\beta}{2}}$

$=\frac{2 \cos \frac{1}{2} \alpha+\frac{\beta}{2}+k \beta+\alpha-\frac{\beta}{2} \sin \frac{1}{2} \alpha+\frac{\beta}{2}+k \beta-\alpha+\frac{\beta}{2}}{2 \sin \frac{\beta}{2}}$

$=\frac{\cos \frac{2 \alpha+k \beta}{2} \sin \frac{k \beta+\beta}{2}}{\sin \frac{\beta}{2}}=\frac{\cos \alpha+\frac{k \beta}{2} \sin (k+1) \frac{\beta}{2}}{\sin \frac{\beta}{2}}=$ RHS

So, $P(k+1)$ is true. Hence, $P(n)$ is true.

Solution

Let $P(n): \cos \theta \cos 2 \theta \ldots \cos 2^{n-1} \theta=\frac{\sin 2^{n} \theta}{2^{n} \sin \theta}$

Step I For $n=1, P(1): \cos \theta=\frac{\sin 2^{1} \theta}{2^{1} \sin \theta}$

$\qquad \qquad \qquad \qquad \qquad \qquad =\frac{\sin 2 \theta}{2 \sin \theta}=\frac{2 \sin \theta \cos \theta}{2 \sin \theta}=\cos \theta $

which is true.

Step II Assume that $P(n)$ is true, for $n=k$.

$ P(k): \cos \theta \cdot \cos 2 \theta \cdot \cos 2^{2} \theta \ldots \cos 2^{k-1} \theta=\frac{\sin 2^{k} \theta}{2^{k} \sin \theta} \text { is true. } $

Step III To prove $P(k+1)$ is true.

$ \begin{aligned} P(k+1): \cos \theta \cdot & \cos 2 \theta \cdot \cos 2^{2} \theta \ldots \cos 2^{k-1} \theta \cdot \cos 2^{k} \theta \\ & =\frac{\sin 2^{k} \theta}{2^{k} \sin \theta} \cdot \cos 2^{k} \theta \\ & =\frac{2 \sin 2^{k} \theta \cdot \cos 2^{k} \theta}{2 \cdot 2^{k} \sin \theta} \\ & =\frac{\sin 2 \cdot 2^{k} \theta}{2^{k+1} \sin \theta}=\frac{\sin 2^{(k+1)} \theta}{2^{k+1} \sin \theta} \end{aligned} $

which is true.

So, $P(k+1)$ is true. Hence, $P(n)$ is true.

Solution

Consider the given statement

$\quad$$P(n): \sin \theta+\sin 2 \theta+\sin 3 \theta+\ldots+\sin n \theta$

$\qquad \qquad =\frac{\sin \frac{n \theta}{2} \sin \frac{(n+1) \theta}{2}}{\sin \frac{\theta}{2}} \text {, for all } n \in N $

Step I We observe that $P(1)$ is

$ \begin{aligned} P(1): \sin \theta & =\frac{\sin \frac{\theta}{2} \cdot \sin \frac{(1+1)}{2} \theta}{\sin \frac{\theta}{2}}=\frac{\sin \frac{\theta}{2} \cdot \sin \theta}{\sin \frac{\theta}{2}} \\ \sin \theta & =\sin \theta \end{aligned} $

Hence, $P(1)$ is true.

Step II Assume that $P(n)$ is true, for $n=k$.

$P(k): \sin \theta+\sin 2 \theta+\sin 3 \theta+\ldots+\sin k \theta$

$ =\frac{\sin \frac{k \theta}{2} \sin \frac{k+1}{2} \theta}{\sin \frac{\theta}{2}} \text { is true. } $

Step III Now, to prove $P(k+1)$ is true.

$P(k+1): \sin \theta+\sin 2 \theta+\sin 3 \theta+\ldots+\sin k \theta+\sin (k+1) \theta$

$ =\frac{\sin \frac{(k+1) \theta}{2} \sin \frac{k+1+1}{2} \theta}{\sin \frac{\theta}{2}} $

LHS $=\sin \theta+\sin 2 \theta+\sin 3 \theta+\ldots+\sin k \theta+\sin (k+1) \theta$

$=\frac{\sin \frac{k \theta}{2} \sin \frac{k+1}{2} \theta}{\sin \frac{\theta}{2}}+\sin (k+1) \theta=\frac{\sin \frac{k \theta}{2} \sin \frac{k+1}{2} \theta+\sin (k+1) \theta \cdot \sin \frac{\theta}{2}}{\sin \frac{\theta}{2}}$ $=\frac{\cos \frac{k \theta}{2}-\frac{k+1}{2} \theta-\cos \frac{k \theta}{2}+\frac{k+1}{2} \theta+\cos (k+1) \theta-\frac{\theta}{2}-\cos (k+1) \theta+\frac{\theta}{2}}{2 \sin \frac{\theta}{2}}$

$=\frac{\cos \frac{\theta}{2}-\cos k \theta+\frac{\theta}{2}+\cos k \theta+\frac{\theta}{2}-\cos k \theta+\frac{3 \theta}{2}}{2 \sin \frac{\theta}{2}}$

$=\frac{\cos \frac{\theta}{2}-\cos k \theta+\frac{3 \theta}{2}}{2 \sin \frac{\theta}{2}}=\frac{2 \sin \frac{1}{2} \frac{\theta}{2}+k \theta+\frac{3 \theta}{2} \cdot \sin \frac{1}{2} k \theta+\frac{3 \theta}{2}-\frac{\theta}{2}}{2 \sin \frac{\theta}{2}}$

$=\frac{\sin \frac{k \theta+2 \theta}{2} \cdot \sin \frac{k \theta+\theta}{2}}{\sin \frac{\theta}{2}}=\frac{\sin (k+1) \frac{\theta}{2} \cdot \sin (k+1+1) \frac{\theta}{2}}{\sin \frac{\theta}{2}}$

So, $P(k+1)$ is true, whenever $P(k)$ is true. Hence, $P(n)$ is true.

Solution

Consider the given statement

$P(n): \frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7 n}{15}$ is a natural number, for all $n \in N$.

Step I We observe that $P(1)$ is true.

$P(1): \frac{(1)^{5}}{5}+\frac{1^{3}}{3}+\frac{7(1)}{15}=\frac{3+5+7}{15}=\frac{15}{15}=1$, which is a natural number. Hence, $P(1)$ is true.

Step II Assume that $P(n)$ is true, for $n=k$.

$P(k): \frac{k^{5}}{5}+\frac{k^{3}}{3}+\frac{7 k}{15}$ is natural number.

Step III Now, to prove $P(k+1)$ is true.

$ \begin{aligned} & \frac{(k+1)^{5}}{5}+\frac{(k+1)^{3}}{3}+\frac{7(k+1)}{15} \\ & =\frac{k^{5}+5 k^{4}+10 k^{3}+10 k^{2}+5 k+1}{5}+\frac{k^{3}+1+3 k(k+1)}{3}+\frac{7 k+7}{15} \\ & =\frac{k^{5}+5 k^{4}+10 k^{3}+10 k^{2}+5 k+1}{5}+\frac{k^{3}+1+3 k^{2}+3 k}{3}+\frac{7 k+7}{15} \\ & =\frac{k^{5}}{5}+\frac{k^{3}}{3}+\frac{7 k}{15}+\frac{5 k^{4}+10 k^{3}+10 k^{2}+5 k+1}{5}+\frac{3 k^{2}+3 k+1}{3}+\frac{7 k+7}{15} \\ & =\frac{k^{5}}{5}+\frac{k^{3}}{3}+\frac{7 k}{15}+k^{4}+2 k^{3}+2 k^{2}+k+k^{2}+k+\frac{1}{5}+\frac{1}{3}+\frac{7}{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 { which is a natural number } \end{aligned} $

So, $P(k+1)$ is true, whenever $P(k)$ is true. Hence,$P(n)$ is true.

Solution

Consider the given statement

$P(n): \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}>\frac{13}{24}$, for all natural numbers $n>1$.

Step I We observe that, $P(2)$ is true,

$ \begin{aligned} P(2): \frac{1}{2+1}+\frac{1}{2+2} & >\frac{13}{24} \\ \frac{1}{3}+\frac{1}{4} & >\frac{13}{24} \\ \frac{4+3}{12} & >\frac{13}{24} \\ \frac{7}{12} & >\frac{13}{24}, \text { which is true. } \end{aligned} $

Hence, $P(2)$ is true.

Step II Now, we assume that $P(n)$ is true,

For $n=k$,

$ P(k): \frac{1}{k+1}+\frac{1}{k+2}+\ldots+\frac{1}{2 k}>\frac{13}{24} $

Step III Now, to prove $P(k+1)$ is true, we have to show that

$ \begin{aligned} & P(k+1): \frac{1}{k+1}+\frac{1}{k+2}+\ldots+\frac{1}{2 k}+\frac{1}{2(k+1)}>\frac{13}{24} \\ & \text { Given, } \\ & \begin{aligned} \frac{1}{k+1}+\frac{1}{k+2}+\ldots+\frac{1}{2 k} & >\frac{13}{24} \\ \because \quad \frac{1}{k+1}+\frac{1}{2 k}+\frac{1}{2(k+1)} & >\frac{13}{24}+\frac{1}{2(k+1)} \\ \frac{13}{24}+\frac{1}{2(k+1)} & >\frac{13}{24} \\ \frac{1}{k+2}+\ldots+\frac{1}{2 k}+\frac{1}{2(k+1)} & >\frac{13}{24} \end{aligned} \end{aligned} $

So, $P(k+1)$ is true, whenever $P(k)$ is true. Hence, $P(n)$ is true.

Solution

Let $P(n)$ : Number of subset of a set containing $n$ distinct elements is $2^{n}$, for all $n \in N$.

Step I We observe that $P(1)$ is true, for $n=1$.

Number of subsets of a set contain 1 element is $2^{1}=2$, which is true.

Step II Assume that $P(n)$ is true for $n=k$.

$P(k)$ : Number of subsets of a set containing $k$ distinct elements is $2^{k}$, which is true.

Step III To prove $P(k+1)$ is true, we have to show that

$P(k+1)$ : Number of subsets of a set containing $(k+1)$ distinct elements is $2^{k+1}$.

We know that, with the addition of one element in the set, the number of subsets become double.

$\therefore$ Number of subsets of a set containing $(k+1)$ distinct elements $=2 \times 2^{k}=2^{k+1}$.

So, $P(k+1)$ is true. Hence, $P(n)$ is true.

Solution

(a) Let $P(n): 10^{n}+3 \cdot 4^{n+2}+k$ is divisible by 9 , for all $n \in N$.

For $n=1$, the given statement is also true $10^{1}+3 \cdot 4^{1+2}+k$ is divisible by 9 .

$\begin{aligned} \because \quad & =10+3 \cdot 64+k=10+192+k \\ & =202+k\end{aligned}$

If $(202+k)$ is divisible by 9 , then the least value of $k$ must be 5 .

$\because \quad 202+5=207$ is divisible by 9

$\Rightarrow \quad \frac{207}{9}=23$

Hence, the least value of $k$ is 5 .

Solution

$(b, c)$

Given that, $3 \cdot 5^{2 n+1}+2^{3 n+1}$

For $n=1$,

Now, $ \begin{aligned} & 3 \cdot 5^{2(1)+1}+2^{3(1)+1} \\ & =3 \cdot 5^{3}+2^{4} \\ & =3 \times 125+16=375+16=391 \\ \qquad \qquad \qquad 391 & =17 \times 23 \end{aligned} $

which is divisible by both 17 and 23 .

Solution

Let $P(n): x^{n}-1$ is divisible by $(x-k)$.

For $n=1, x^{1}-1$ is divisible by $(x-k)$.

Since, if $x-1$ is divisible by $x-k$. Then, the least possible integral value of $k$ is 1 .

Solution

Given that, $P(n): 2 n<n !, n \in N$

For $n=1, \qquad$ $2<$ !$\quad$ [false]

For $n=2, \qquad 2 \times 2<2 ! 4<2$ $\quad$ [false]

For $n=3, \qquad 2 \times 3<3$ ! $6<3$ !

$\qquad \qquad \qquad \quad 6<3 \times 2 \times 1$

$\qquad \qquad \qquad \quad (6<6) \qquad$ [false]

For $n=4$, $\qquad 2 \times 4<4$ !

$\qquad \qquad \qquad \quad 8<4 \times 3 \times 2 \times 1$ $(8<24)$ $\qquad$ [true ]

For $n=5$, $\qquad 2 \times 5<5$ !

$\qquad \qquad \qquad 10<5 \times 4 \times 3 \times 2 \times 1$

$\qquad \qquad \qquad (10<120)$ $\qquad$ $\quad$ [true]

Hence, $P(n)$ is for all $n \geq 4$.

True/False

Solution

False

The given statement is false because $P(1)$ is true has not been proved.