Solution

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

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

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

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

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

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

Solution

Consider the statement

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

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.

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 Sol.ve 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

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

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

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

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 $P(n):2^{3n}-1$ 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 $n=k$, $P(k):2^{3k}-1$ is divisible by 7 .

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

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

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

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

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

Hence, $P(1)$ is true.

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

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

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

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

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

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

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

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

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

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

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

$\begin{array}{rl}P(k+1)& :3^{2(k+1)}-1\\ & =3^{2k}\cdot 3^2-1\\ & =3^{2k}\cdot (8+1)-1\\ & =8\cdot 3^{2k}+3^{2k}-1\\ & =8\cdot 3^{2k}+8q\\ & =8(3^{2k}+q)\end{array}$ $$ [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 $n=k$.

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

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

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

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

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

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 $x\ne y$.

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 $n=k$.

$P(k):x^k-y^k$ is divisible by (x-y) .

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

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

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

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 $n\ge 2$.

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

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

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

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

$\therefore k^3-k=6q$

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

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

We know that, $3k(k+1)$ 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 .

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

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

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

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

Since, $P(k+1):6q+3(k^2+k+2)$ 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 $n\ge 5$.

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

Hence, $P(5)$ is true.

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

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,

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

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

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 $n\ge 5$.

Solution

Consider the statement

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

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

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

Hence, $P(1)$ is true.

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

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

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

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

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

$$ 2 k <(k+2) !

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

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

$$ (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

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

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 $n=k$.

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

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

Given that,

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

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

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

If

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

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

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

From Eqs. (i) and (ii),

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

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

Solution

Let $P(n):2+4+6+\dots +2n=n^2+n$

For all natural numbers $n$.

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

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

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

$\therefore P(k):2+4+6+\dots +2k=k^2+k$

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

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

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

Hence, $P(n)$ is true.

Solution

Consider the given statement

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

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

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

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

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

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

$\begin{array}{rl}P(k+1):& 1+2+2^2+\dots +2^k+2^{k+1}\\ & =2^{k+1}-1+2^{k+1}\\ & =2\cdot 2^{k+1}-1\\ & =2^{k+2}-1\\ & =2^{(k+1)+1}-1\end{array}$

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

Hence, $P(n)$ is true.

Solution

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

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

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

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 $a_1,a_2,a_3,\dots$ is defined by letting $a_1=3$ and $a_k=7a_{k-1}$, for all natural numbers $k\ge 2$.

Let $P(n):a_n=3\cdot 7^{n-1}$ 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

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

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

Solution

Consider the given statement,

$P(n):b_n=5+4n$, 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 $b_0=5,b_1=4+b_0=4+5=9$

Hence, $P(1)$ is true.

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

$P(k):b_k=5+4k$

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

$\begin{array}{rl}& \therefore P(k+1):b_{k+1}=5+4(k+1)\\ & b_{k+1}=4+b_{k+1-1}\\ & =4+b_k\\ & =4+5+4k=5+4(k+1)\end{array}$

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

Solution

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

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

As, given $d_1=2$

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

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

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 $P(k+1):d_{k+1}=\frac{2}{(k+1)!}$

$\begin{array}{rl}{d}_{k+1}& =\frac{d_{k+1-1}}{k}=\frac{d_k}{k}\\ & =\frac{2}{k!k}=\frac{2}{(k+1)!}\end{array}$

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

$\sin A-\sin B=2\cos \frac{A+B}{2}\cdot \sin \frac{A-B}{2}$

Solution

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

Step I We observe that $P(1)$

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

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

Hence, $P(1)$ is true.

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

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

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

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

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

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

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

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

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

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

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

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

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

Solution

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

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

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

which is true.

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

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

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

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

which is true.

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

Thinking Process

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

$$ $\cos A-\cos B=2\sin \frac{A+B}{2}\cdot \sin \frac{B-A}{2}$ also $\cos (-\theta )=\cos \theta$.

Solution

Consider the given statement

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

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

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

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

Hence, $P(1)$ is true.

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

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

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

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

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

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

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

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

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

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