Solution

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

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

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

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

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

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

Solution

Consider the statement

$P(n): 1^{2}+2^{2}+3^{2}+\ldots+n^{2} =\dfrac{n(n+1)(2 n+1)}{6} $

$\text { For } \ n \ =1, =\dfrac{1(1+1)(2 \times 1+1)}{6} $

$\Rightarrow\quad 1 =\dfrac{2(3)}{6} $

$\Rightarrow\quad 1=1 $

$\text { For } \ n \ =2, 1+2^{2} =\dfrac{2(2+1)(4+1)}{6} $

$\Rightarrow\quad 5=\dfrac{30}{6} $

$\Rightarrow\quad 5=5 $

$\text { For } \ n \ =3, 1+2^{2}+3^{2} =\dfrac{3(3+1)(7)}{6}$

$\Rightarrow\quad 1+4+9 =\dfrac{3(4)(7)}{6}$

$\Rightarrow\quad 14 = 14$

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

Thinking Process

In step I put $n=1$, the obtained result should be a divisible by $3$. In step II put $n=k$ and take $P(k)$ equal to multiple of $3$ with non-zero constant say $q$. In step III put $n=k+1$, in the statement and Solve till it becomes a multiple of $3.$

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 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) :\quad 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 \quad [\text{from step}\ II] \\ \\ & =8(3^{2 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 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{aligned} P(k+1): 7^{k+1}-2^{k+1} & =7^{k} \cdot 7-2^{k} \cdot 2 \\ \\ & =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}) \text { [from step II] } \\ \\ & =5 \cdot 7^{k}+2(5 q) \end{aligned} $

$ \Rightarrow\quad 5(7^{k}+2 q) \text {, which is divisible by } 5 $

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 $n$.

Thinking Process

In step I put $n=2$, the obtained result should be divisible by 6. Then, follow the same process as in question no. 4.

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) \quad [\text {from step}\ II] \end{aligned} $

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} & \Rightarrow k^{2}+2 k+1<2^{k}+2 k+1 \\ \\ & \Rightarrow (k+1)^{2}<2^{k}+2 k+1 \ldots(i)\\ \\ \end{aligned} $

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

From eq. (i) and (ii), we get

$\therefore$ $(k+1)^{2}<2^{k+1}$

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 \quad 2<3 ! $

$\Rightarrow \quad 2<3 \times 2 \times 1$

$ \Rightarrow\quad 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+2 <(k+2) !+2 $

$\qquad \qquad \qquad 2(k+1) < (k+2) !+2 < (k+3) ! \qquad [\because \ \ (k+2) !+2 <(k+3) !]$

$ \therefore \quad 2(k+1)<(k+3) ! $

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

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

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

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

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

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

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

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

Given that,

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

Adding $\dfrac{1}{\sqrt{k+1}} $ in both sides, we get

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

$\Rightarrow \quad \dfrac{(\sqrt{k})(\sqrt{k+1})+1}{\sqrt{k+1}}<\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}} \ldots(i)$

We need to prove that, $ \sqrt{k+1}<\dfrac{\sqrt{k} \sqrt{k+1}+1}{\sqrt{k+1}} $

$ Now, \quad k+1<\sqrt{k} \sqrt{k+1}+1 $

$ \sqrt{k+1}<\dfrac{\sqrt{k} \sqrt{k+1}+1}{\sqrt{k+1}}\ldots(ii) $

From Eqs. (i) and (ii),

$ \sqrt{k+1}<\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\ldots+\dfrac{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.

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} \\ \\ \text{Now}, \ 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} & \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 $P(n): d_{n}=\dfrac{2}{n !}, \forall n \in N$, to prove $P(2)$ is true.

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

As, given $\quad d_1=2 $

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

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

Hence, $P(2)$ is true.

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

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

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

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

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

Thinking Process

To prove this, use the formula $2 \cos A \sin B=\sin (A+B)-\sin (A-B)$ and

$ \sin A- \sin =2 \cos \left(\dfrac{A+B}{2}\right ) \cdot \sin \left(\dfrac{A-B}{2} \right) $

Solution

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

Step I: We observe that,

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

$ P(1): \cos \alpha=\dfrac{\cos \left(\alpha+\dfrac{1-1}{2} \beta\right) \sin \dfrac{\beta}{2}}{\sin \dfrac{\beta}{2}}=\dfrac{\cos (\alpha+0) \sin \dfrac{\beta}{2}}{\sin \dfrac{\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] \\ \\ &= \dfrac{\cos \left(\alpha+\dfrac{k-1}{2} \beta\right) \sin \dfrac{k \beta}{2}}{\sin \dfrac{\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]=\dfrac{\cos (\alpha+\dfrac{k \beta}{2}) \sin (k+1) \dfrac{\beta}{2}}{\sin \dfrac{\beta}{2}} $

$\text{L.H.S} \ \ =\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\ldots+\cos [\alpha+(k-1) \beta]+\cos (\alpha+k \beta)$

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

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

$\qquad \ \quad=\dfrac{\sin \left(\alpha+\dfrac{k \beta}{2}-\dfrac{\beta}{2}+\dfrac{k \beta}{2}\right)-\sin \left(\alpha+\dfrac{k \beta}{2}-\dfrac{\beta}{2}-\dfrac{k \beta}{2}\right)+\sin \left(\alpha+k \beta+\dfrac{\beta}{2}\right)-\sin \left(\alpha+k \beta-\dfrac{\beta}{2}\right)}{2 \sin \dfrac{\beta}{2}}$

$\qquad \ \quad=\dfrac{\sin \left(\alpha+k \beta+\dfrac{\beta}{2}\right)-\sin \left(\alpha-\dfrac{\beta}{2}\right)}{2 \sin \dfrac{\beta}{2}}$

$\qquad \ \quad=\dfrac{2 \cos \left(\dfrac{1}{2} \alpha+\dfrac{\beta}{2}+k \beta+\alpha-\dfrac{\beta}{2}\right) \sin \left(\dfrac{1}{2} \alpha+\dfrac{\beta}{2}+k \beta-\alpha+\dfrac{\beta}{2}\right)}{2 \sin \dfrac{\beta}{2}}$

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

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

Solution

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

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

$\qquad \qquad \qquad \qquad \qquad \qquad =\dfrac{\sin 2 \theta}{2 \sin \theta}=\dfrac{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=\dfrac{\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 \\ \\ & =\dfrac{\sin 2^{k} \theta}{2^{k} \sin \theta} \cdot \cos 2^{k} \theta \\ \\ & =\dfrac{2 \sin 2^{k} \theta \cdot \cos 2^{k} \theta}{2 \cdot 2^{k} \sin \theta} \\ \\ & =\dfrac{\sin 2 \cdot 2^{k} \theta}{2^{k+1} \sin \theta}=\dfrac{\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 =\dfrac{\sin \dfrac{n \theta}{2} \sin \dfrac{(n+1) \theta}{2}}{\sin \dfrac{\theta}{2}} \text {, for all } n \in N $

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

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

Hence, $P(1)$ is true.

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

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

$ =\dfrac{\sin \dfrac{k \theta}{2} \sin \dfrac{(k+1)}{2} \theta}{\sin \dfrac{\theta}{2}} \text { is true. } $

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

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

$ \hspace{1cm}=\dfrac{\sin \dfrac{(k+1) \theta}{2} \sin \dfrac{(k+1+1)}{2} \theta}{\sin \dfrac{\theta}{2}} $

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

$\hspace{1cm}=\dfrac{\sin \dfrac{k \theta}{2} \sin \dfrac{(k+1)}{2} \theta}{\sin \dfrac{\theta}{2}}+\sin (k+1) \theta=\dfrac{\sin \dfrac{k \theta}{2} \sin \dfrac{(k+1)}{2} \theta+\sin (k+1) \theta \cdot \sin \dfrac{\theta}{2}}{\sin \dfrac{\theta}{2}}$

$\hspace{1cm}=\dfrac{\cos \left[\dfrac{k \theta}{2}-\dfrac{(k+1)}{2} \theta\right]-\cos \left[\dfrac{k \theta}{2}+\dfrac{(k+1)}{2} \theta\right] +\cos \left[(k+1) \theta-\dfrac{\theta}{2}\right]-\cos\left[ (k+1) \theta+\dfrac{\theta}{2}\right]}{2 \sin \dfrac{\theta}{2}}$

$\hspace{1cm}=\dfrac{\cos \dfrac{\theta}{2}-\cos k \theta+\dfrac{\theta}{2}+\cos k \theta+\dfrac{\theta}{2}-\cos k \theta+\dfrac{3 \theta}{2}}{2 \sin \dfrac{\theta}{2}}$

$\hspace{1cm}=\dfrac{\cos \dfrac{\theta}{2}-\cos\left( k \theta+\dfrac{3 \theta}{2}\right)}{2 \sin \dfrac{\theta}{2}}=\dfrac{2 \sin \left[\dfrac{1}{2} \left(\dfrac{\theta}{2}+k \theta+\dfrac{3 \theta}{2}\right]\right) \cdot \sin \left[\dfrac{1}{2} \left(k \theta+\dfrac{3 \theta}{2}-\dfrac{\theta}{2}\right)\right]}{2 \sin \dfrac{\theta}{2}}$

$\hspace{1cm}=\dfrac{\sin \left(\dfrac{k \theta+2 \theta}{2}\right) \cdot \sin \left(\dfrac{k \theta+\theta}{2}\right)}{\sin \dfrac{\theta}{2}}=\dfrac{\sin \left(k+1\right) \dfrac{\theta}{2} \cdot \sin \left(k+1+1\right) \dfrac{\theta}{2}}{\sin \dfrac{\theta}{2}}$

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

Thinking Process

Here, use the formula $(a+b)^{5}=a^{5}+5 a b^{4}+10 a^{2} b^{3}+10 a^{3} b^{2}+5 a^{4} b+b^{5}$

and $\quad (a+b)^{3}=a^{3}+b^{3}+3 a b(a+b) $

Solution

Consider the given statement

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

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

$P(1): \dfrac{(1)^{5}}{5}+\dfrac{1^{3}}{3}+\dfrac{7(1)}{15}=\dfrac{3+5+7}{15}=\dfrac{15}{15}=1$,

which is a natural number. Hence, $P(1)$ is true.

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

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

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

$ \begin{aligned} & \dfrac{(k+1)^{5}}{5}+\dfrac{(k+1)^{3}}{3}+\dfrac{7(k+1)}{15} \\ \\ & =\dfrac{k^{5}+5 k^{4}+10 k^{3}+10 k^{2}+5 k+1}{5}+\dfrac{k^{3}+1+3 k(k+1)}{3}+\dfrac{7 k+7}{15} \\ \\ & =\dfrac{k^{5}+5 k^{4}+10 k^{3}+10 k^{2}+5 k+1}{5}+\dfrac{k^{3}+1+3 k^{2}+3 k}{3}+\dfrac{7 k+7}{15} \\ \\ & =\dfrac{k^{5}}{5}+\dfrac{k^{3}}{3}+\dfrac{7 k}{15}+\dfrac{5 k^{4}+10 k^{3}+10 k^{2}+5 k+1}{5}+\dfrac{3 k^{2}+3 k+1}{3}+\dfrac{7 k+7}{15} \\ \\ & =\dfrac{k^{5}}{5}+\dfrac{k^{3}}{3}+\dfrac{7 k}{15}+k^{4}+2 k^{3}+2 k^{2}+k+k^{2}+k+\dfrac{1}{5}+\dfrac{1}{3}+\dfrac{7}{15} \\ \\ & =\dfrac{k^{5}}{5}+\dfrac{k^{3}}{3}+\dfrac{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

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

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

$ \begin{aligned} P(2): \dfrac{1}{2+1}+\dfrac{1}{2+2} & >\dfrac{13}{24} \\ \\ \dfrac{1}{3}+\dfrac{1}{4} & >\dfrac{13}{24} \\ \\ \dfrac{4+3}{12} & >\dfrac{13}{24} \\ \\ \dfrac{7}{12} & >\dfrac{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): \dfrac{1}{k+1}+\dfrac{1}{k+2}+\ldots+\dfrac{1}{2 k}>\dfrac{13}{24} $

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

$ \begin{aligned} & P(k+1): \dfrac{1}{k+1}+\dfrac{1}{k+2}+\ldots+\dfrac{1}{2 k}+\dfrac{1}{2(k+1)}>\dfrac{13}{24} \\ \\ & \text { Given, } \\ \\ & \begin{aligned} \dfrac{1}{k+1}+\dfrac{1}{k+2}+\ldots+\dfrac{1}{2 k} & >\dfrac{13}{24} \\ \\ \because \quad \dfrac{1}{k+1}+\dfrac{1}{2 k}+\dfrac{1}{2(k+1)} & >\dfrac{13}{24}+\dfrac{1}{2(k+1)} \\ \\ \dfrac{13}{24}+\dfrac{1}{2(k+1)} & >\dfrac{13}{24} \\ \\ \dfrac{1}{k+2}+\ldots+\dfrac{1}{2 k}+\dfrac{1}{2(k+1)} & >\dfrac{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 $9 .$

$\begin{aligned} \because \quad 10^{1}+3 \cdot 4^{1+2}+k & =10+3 \cdot 64+k=10+192+k \\ \\ & =202+k\end{aligned}$

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

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

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

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,

$ 3 \cdot 5^{2(1)+1}+2^{3(1)+1} =3 \cdot 5^{3}+2^{4}$

$ \qquad \qquad \qquad \qquad=3 \times 125+16$

$\qquad \qquad\qquad \qquad=375+16=391$

$\qquad \qquad \qquad 391 =17 \times 23$

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

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

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

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

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

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

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

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

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

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

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

True/False

Solution

True

Given, $P(n) \Rightarrow P(n+1)$

replace, $n$ with $n-1$

$P(n-1) \Rightarrow P(n-1+1) $

$P(n-1) \Rightarrow P(n) $

$\Rightarrow \quad P(n)$ is true for both $(n-1)$ and $(n+1)$

It is true foll all $n \in N$.