Important Questions for CBSE Class 11 Maths Chapter 4 - Principle of Mathematical Induction

4 Marks Questions

1.     For every integer n, prove that $\mathbf{{{7}^{n}}-{{3}^{n}}}$ is divisible by 4.

Ans: In this Question we have to prove that ${{7}^{n}}-{{3}^{n}}$ is divisible by \[4\] by using the method of PMI.

Given, \[P\text{ }\left( n \right):\]${{7}^{n}}-{{3}^{n}}$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] ${{7}^{1}}-{{3}^{1}}=4$

Which is divisible by \[4\] .

Thus,  \[P\text{ }\left( 1 \right)\] is true.

Let,  \[P\text{ }\left( n \right):\]${{7}^{n}}-{{3}^{n}}$ is true for $n=k$ .

That is,  \[P\text{ }\left( k \right):\] ${{7}^{k}}-{{3}^{k}}=4\lambda \text{ },where\text{ }\lambda \in \text{N    }.......\text{(i)}$

Now, we have to show that the given statement  \[P\text{ }\left( n \right):\]${{7}^{n}}-{{3}^{n}}$ is true for $n=k+1$ .

\[P\text{ }\left( k+1 \right):\] ${{7}^{k+1}}-{{3}^{k+1}}={{7}^{k}}\cdot 7-{{3}^{k}}\cdot 3$

$(4 \lambda + 3^k) \cdot 7 - 3^k \cdot 3 $ from equation (1)

$\text{28}\lambda \text{+7}\cdot \text{3}-{{3}^{k}}\cdot 3$

$28\lambda +{{3}^{k}}(7-3)$

$4(7\lambda +{{3}^{k}})$

Hence, \[P\text{ }\left( k+1 \right):\] ${{7}^{k+1}}-{{3}^{k+1}}$ is divisible by  \[4.\]

Thus, \[P\text{ }\left( k+1 \right)\] is true \[when\text{ }P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the statement is true for every positive integer \[n\] .

2.     Prove that n(n+1)(n+5) is multiple of 3.

Ans: In this Question we have to prove that $n(n+1)(n+5)$ is multiple of \[3\] by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $n(n+1)(n+5)$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $1(1+1)(1+5)=12$

Which is multiple of \[3\].

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $n(n+1)(n+5)$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] $k(k+1)(k+5)=3\lambda \text{ },where\text{ }\lambda \in \text{N    }.......\text{(i)}$

Now, we have to show that the given statement \[P\text{ }\left( n \right):\] $n(n+1)(n+5)$ is true for $n=k+1$.

\[P\text{ }\left( k+1 \right):\]$(k+1)(k+2)(k+6)$

\[(k+1)(k+2)(k+6)=[(k+1)(k+2)](k+6)\]

$k(k+1)(k+2)+6(k+1)(k+2)$

$k(k+1)(k+5-3)+6(k+1)(k+2)$

$k(k+1)(k+5)-3k(k+1)+6(k+1)(k+2)$

$k(k+1)(k+5)+(k+1)[6(k+2)-3k]$

$k(k+1)(k+5)+(k+1)(3k+12)$

$k(k+1)(k+5)+3(k+1)(k+4)$

$3\lambda +3(k+1)(k+4) $ from equation (1)

$3[\lambda \text{ }+(k+1)(k+4)]$

Hence,  \[P\text{ }\left( k+1 \right):\] $(k+1)(k+2)(k+6)$  is multiple of \[3\] .

Thus,  \[P\text{ }\left( k+1 \right)\]is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the given statement $n\cdot (n+1)\cdot (n+5)$ is multiple of \[~3\].

3.     Prove that  $\mathbf{{{10}^{2n-1}}+1}$  is divisible by \[\mathbf{11}\].

Ans: In this Question we have to prove that ${{10}^{2n-1}}+1$ is divisible by  \[11\]  by using the method of PMI.

Given, \[P\text{ }\left( n \right):\]${{10}^{2n-1}}+1$.

Checking if the statement is true or not for $n=1$.

So, For $n=1$

\[P\text{ }\left( 1 \right):\] ${{10}^{(2\cdot 1)-1}}+1=11$

Which is divisible by \[11\].

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\]${{10}^{2n-1}}+1$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] ${{10}^{2n-1}}+1=11k\text{ },where\text{ }\lambda \in \text{N    }.......\text{(i)}$

Now, we have to show that the given statement \[P\text{ }\left( n \right):\] ${{10}^{2n-1}}+1$ is true for $n=k+1$.

\[P\text{ }\left( k+1 \right):\] ${{10}^{2(k+1)-1}}+1={{10}^{2k+2-1}}+1$

${{10}^{2k+1}}+1$

$({{10}^{2k}}\cdot 10)+1$

$(110 \lambda - 10) \cdot 10+1 $ from equation (i)

$1100\lambda -100+1$

$1100\lambda -99$

$11(100\lambda -99)$

Hence, \[P\text{ }\left( k+1 \right):\] ${{10}^{2(k+1)-1}}+1$ is divisible by \[11\].

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the statement ${{10}^{2n-1}}+1$ is multiple of \[11\] .

4.     Prove $\mathbf{\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{n} \right)=(n+1)}$ .

Ans: In this Question we have to prove that $\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{n} \right)=(n+1)$ by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{n} \right)=(n+1)$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $\left( 1+\dfrac{1}{1} \right)=(1+1)$

$\left( 1+\dfrac{1}{1} \right)=2$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{n} \right)=(n+1)$is true for $n=k$.

That is, \[P\text{ }\left( k \right):\] $\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{k} \right)=(k+1) $ $\text{.......(i)}$

Now, we have to show that the given statement \[~P\text{ }\left( n \right):\] $\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{n} \right)=(n+1)$ is true for $n=k+1$ .

\[P\text{ }\left( k+1 \right):\] $\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{n} \right)\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{k} \right)\left( 1+\dfrac{1}{k+1} \right)=(k+1+1)$

$\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{n} \right)\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{k} \right)\left( 1+\dfrac{1}{k+1} \right)=(k+2)$

Now, L.H.S=$\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{n} \right)\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{k} \right)\left( 1+\dfrac{1}{k+1} \right)$

$(k+1)\left( 1+\dfrac{1}{k+1} \right) $ …….. (1)

$(k+1)\left( \dfrac{k+1+1}{k+1} \right)$

$(k+2)=R.H.S$

Hence, \[P\text{ }\left( k+1 \right):\] $\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{n} \right)\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{k} \right)\left( 1+\dfrac{1}{k+1} \right)=(k+2)$  is true.

Thus \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I.

The statement $\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{n} \right)\left( 1+\dfrac{1}{1} \right)\left( 1+\dfrac{1}{2} \right)\left( 1+\dfrac{1}{3} \right)......\left( 1+\dfrac{1}{k} \right)\left( 1+\dfrac{1}{k+1} \right)=(k+2)$ is true.

5.     Prove $\mathbf{1\cdot 2+2\cdot 3+3\cdot 4+.......+n(n+1)=\dfrac{n(n+1)(n+2)}{3}}$.

Ans: In this Question we have to prove that $1\cdot 2+2\cdot 3+3\cdot 4+.......+n(n+1)=\dfrac{n(n+1)(n+2)}{3}$ by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $1\cdot 2+2\cdot 3+3\cdot 4+.......+n(n+1)=\dfrac{n(n+1)(n+2)}{3}$.

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $1\cdot 2=\dfrac{1(1+1)(1+2)}{3}$

$2=\dfrac{6}{3}$

$2=2$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $1\cdot 2+2\cdot 3+3\cdot 4+.......+n(n+1)=\dfrac{n(n+1)(n+2)}{3}$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] $1\cdot 2+2\cdot 3+3\cdot 4+.......+k(k+1)=\dfrac{k(k+1)(k+2)}{3}\text{      }...........\text{(i)}$

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $1\cdot 2+2\cdot 3+3\cdot 4+.......+n(n+1)=\dfrac{n(n+1)(n+2)}{3}$  Is true for $n=k+1$ .

So, \[P\text{ }\left( k+1 \right):\] $1\cdot 2+2\cdot 3+3\cdot 4+.......+(k+1)(k+2)=\dfrac{(k+1)(k+2)(k+3)}{3}$

Now, L.H.S  $1\cdot 2+2\cdot 3+3\cdot 4+.......+k(k+1)+(k+1)(k+2)$

$\dfrac{k(k+1)(k+2)}{3}\text{+}\dfrac{(k+1)(k+2)}{1}$ from equation (1)

$\dfrac{k(k+1)(k+2)+3(k+1)(k+2)}{3}$

$\dfrac{(k+3)(k+1)(k+2)}{3}=R.H.S$

Hence, \[P\text{ }\left( k+1 \right):\] $1\cdot 2+2\cdot 3+3\cdot 4+.......+(k+1)(k+2)=\dfrac{(k+1)(k+2)(k+3)}{3}$ is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I.

The statement $1\cdot 2+2\cdot 3+3\cdot 4+.......+n(n+1)=\dfrac{n(n+1)(n+2)}{3}$ is true.

6.     Prove \[\mathbf{\left( \mathbf{2n}+\mathbf{7} \right)<{{\left( \mathbf{n}+\mathbf{3} \right)}^{2}}}\]

Ans: In this Question we have to prove that \[\left( \mathbf{2n}+\mathbf{7} \right)<{{\left( \mathbf{n}+\mathbf{3} \right)}^{2}}\] by using the method of P.M.I.

Given, \[P\text{ }\left( n \right):\] \[\left( \mathbf{2n}+\mathbf{7} \right)<{{\left( \mathbf{n}+\mathbf{3} \right)}^{2}}\] .

Checking if the statement is true or not for $n=1$ .

So, For$n=1$

\[P\text{ }\left( 1 \right):\] $9<{{(4)}^{2}}$

$9<16$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] \[\left( \mathbf{2n}+\mathbf{7} \right)<{{\left( \mathbf{n}+\mathbf{3} \right)}^{2}}\] is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] \[\left( \mathbf{2}k+\mathbf{7} \right)<{{\left( k+\mathbf{3} \right)}^{2}}\text{        }...........\text{(i)}\]

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] \[[2(k+1)+7]<{{\left( k+1+\mathbf{3} \right)}^{2}}\] Is true for $n=k+1$.

So, \[P\text{ }\left( k+1 \right):\] \[(2k+9)<{{\left( k+4 \right)}^{2}}\]

Now, \[(2k+7+2)<{{\left( k+4 \right)}^{2}}\]

${{(k+3)}^{2}}+2<{{\left( k+4 \right)}^{2}}$ from equation (1)

${{k}^{2}}+9+6k+2<{{(k+4)}^{2}}$

${{k}^{2}}+6k+11<{{(k+4)}^{2}}$

${{k}^{2}}+6k+11+2k-2k+5-5<{{(k+4)}^{2}}$

$({{k}^{2}}+8k+16)-(2k+5)<{{(k+4)}^{2}}$

${{(k+4)}^{2}}-(2k+5)<{{(k+4)}^{2}}$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] \[(2k+9)<{{\left( k+4 \right)}^{2}}\] is true.

Thus \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the statement \[\left( \mathbf{2n}+\mathbf{7} \right)<{{\left( \mathbf{n}+\mathbf{3} \right)}^{2}}\] is true for all  n $\in $ N.

7.     Prove $\mathbf{\dfrac{1}{1\cdot 4}+\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+......+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{(3n+1)}}$ .

Ans: In this Question we have to prove that$\dfrac{1}{1\cdot 4}+\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+......+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{(3n+1)}$ by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $\dfrac{1}{1\cdot 4}+\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+......+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{(3n+1)}$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $\dfrac{1}{1\cdot 4}=\dfrac{1}{(3\cdot 1)+1}$

$\dfrac{1}{4}=\dfrac{1}{4}$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $\dfrac{1}{1\cdot 4}+\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+......+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{(3n+1)}$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] $\dfrac{1}{1\cdot 4}+\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+......+\dfrac{1}{(3k-2)(3k+1)}=\dfrac{k}{(3k+1)}$ —- (i)

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $\dfrac{1}{1\cdot 4}+\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+......+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{(3n+1)}$  Is true for $n=k+1$ .

So, \[P\text{ }\left( k+1 \right):\] $\dfrac{1}{1\cdot 4}+\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+......+\dfrac{1}{[3(k+1)-2][3(k+1)+1]}=\dfrac{k+1}{3(k+1)+1}$ .

Now, L.H.S $=\dfrac{1}{1\cdot 4}+\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+......+\dfrac{1}{(3k-2)(3k+1)}+\dfrac{1}{[3(k+1)-2][3(k+1)+1]}$

$=\dfrac{k}{3k+1}\text{+}\dfrac{1}{(3k+1)(3k+4)} $ from equation (1)

$=\dfrac{k(3k+4)+1}{(3k+1)(3k+4)}$

$=\dfrac{3{{k}^{2}}+4k+1}{(3k+1)(3k+4)}$

$=\dfrac{(3k+1)(k+1)}{(3k+1)(3k+4)}$

$=\dfrac{(k+1)}{(3k+3+1)}$

$= \dfrac{(k+1)}{3(k+1)+1}$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] $\dfrac{1}{1\cdot 4}+\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+......+\dfrac{1}{[3(k+1)-2][3(k+1)+1]}=\dfrac{k+1}{3(k+1)+1}$  is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the statement $\dfrac{1}{1\cdot 4}+\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+......+\dfrac{1}{(3n-2)(3n+1)}=\dfrac{n}{(3n+1)}$ is true.

8.     Prove $\mathbf{1\cdot 2+2\cdot {{2}^{2}}+3\cdot {{2}^{3}}+...+n\cdot {{2}^{n}}=(n-1){{2}^{n+1}}+2}$

Ans: In this Question we have to prove that$1\cdot 2+2\cdot {{2}^{2}}+3\cdot {{2}^{3}}+...+n\cdot {{2}^{n}}=(n-1){{2}^{n+1}}+2$ by using the method of PMI.

Given, \[P\text{ }\left( 1 \right):\] $1\cdot 2+2\cdot {{2}^{2}}+3\cdot {{2}^{3}}+...+n\cdot {{2}^{n}}=(n-1){{2}^{n+1}}+2$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] \[1\cdot {{2}^{1}}=(1-1){{2}^{2}}+2\]

$2=2$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $1\cdot 2+2\cdot {{2}^{2}}+3\cdot {{2}^{3}}+...+n\cdot {{2}^{n}}=(n-1){{2}^{n+1}}+2$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] \[1\cdot 2+2\cdot {{2}^{2}}+3\cdot {{2}^{3}}+...+k\cdot {{2}^{k}}=(k-1){{2}^{k+1}}+2\] —- (i)

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $1\cdot 2+2\cdot {{2}^{2}}+3\cdot {{2}^{3}}+...+n\cdot {{2}^{n}}=(n-1){{2}^{n+1}}+2$ Is true for $n=k+1$.

So, \[P\text{ }\left( k+1 \right):\] $1\cdot 2+2\cdot {{2}^{2}}+3\cdot {{2}^{3}}+...+(k+1)\cdot {{2}^{k+1}}=(k+1-1){{2}^{k+1+1}}+2$

Now, L.H.S $=1\cdot 2+2\cdot {{2}^{2}}+3\cdot {{2}^{3}}+...+(k+1)\cdot {{2}^{k+1}}$

$1 \cdot 2 +  $

$\text{=1}\cdot \text{2+2}\cdot {{\text{2}}^{2}}\text{+}....\text{+k}\cdot {{\text{2}}^{2}}\text{+(k+1)}{{\text{2}}^{k+1}}\text{   }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

$=(k-1){{2}^{k+1}}+2+(k+1){{2}^{K+1}}$

$={{2}^{k+1}}(k-1+k+1)+2$

${{2}^{k+2}}(k+2)=R.H.S$

Hence, \[P\text{ }\left( k+1 \right):\] $1\cdot 2+2\cdot {{2}^{2}}+3\cdot {{2}^{3}}+...+(k+1)\cdot {{2}^{k+1}}=(k){{2}^{k+2}}+2$ is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I.

The statement $1\cdot 2+2\cdot {{2}^{2}}+3\cdot {{2}^{3}}+...+n\cdot {{2}^{n}}=(n-1){{2}^{n+1}}+2$ is true.

9.     Prove that $\mathbf{2\cdot {{7}^{n}}+3\cdot {{5}^{n}}-5}$ is divisible by \[\mathbf{24}\forall n\in N\] .

Ans: In this Question we have to prove that $2\cdot {{7}^{n}}+3\cdot {{5}^{n}}-5$ divisible by $24$ by using the method of PMI.

Given, \[P\text{ }\left( 1 \right):\] $2\cdot {{7}^{n}}+3\cdot {{5}^{n}}-5$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] \[2\cdot {{7}^{1}}+3\cdot {{5}^{1}}-5=24\]

Which is divisible by $24$ .

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $2\cdot {{7}^{n}}+3\cdot {{5}^{n}}-5$ is divisible by $24$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] $2\cdot {{7}^{k}}+3\cdot {{5}^{k}}-5=24\lambda $  where $\lambda \in N$       …….. (i)

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $2\cdot {{7}^{n}}+3\cdot {{5}^{n}}-5$  Is divisible by $24$ for $n=k+1$ .

So, \[P\text{ }\left( k+1 \right):\] $2\cdot {{7}^{k+1}}+3\cdot {{5}^{k+1}}-5$

$=2\cdot {{7}^{k}}\cdot 7+3\cdot {{5}^{k}}\cdot 5-5$

$\text{=7 }\!\![\!\!\text{ 2}\cdot {{\text{7}}^{k}}\text{+3}\cdot {{\text{5}}^{k}}\text{-5-3}\cdot {{\text{5}}^{k}}+5]\text{   }$

$=7[24\lambda -3\cdot {{5}^{k}}\text{+5 }\!\!]\!\!\text{  + 15}\cdot {{\text{5}}^{k}}\text{-5   }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

$=7\cdot 24\lambda -6\cdot {{5}^{k}}+30$

$=7\cdot 24\lambda -6({{5}^{k}}-5)$

$=7\cdot 24\lambda -6\cdot 4p$ since ${{\text{5}}^{k}}-5$ $\text{ is multiple of 4,Let }{{\text{5}}^{k}}-5=p\}$

$=24(7\lambda -p)$

Hence,  \[P\text{ }\left( k+1 \right):\] $2\cdot {{7}^{k+1}}+3\cdot {{5}^{k+1}}-5$ is divisible by $24$ is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I.

The statement $2\cdot {{7}^{n}}+3\cdot {{5}^{n}}-5$ is divisible by \[24\forall n\in N\] is true.

10.  Prove that $\mathbf{{{41}^{n}}-{{14}^{n}}}$  is a multiple of 27.

Ans: In this Question we have to prove that ${{41}^{n}}-{{14}^{n}}$ is multiple of \[27\] by using the method of PMI.

Given,  \[P\text{ }\left( n \right):\] ${{41}^{n}}-{{14}^{n}}$ .

Checking if the statement is true or not for $n=1$.

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $41-14=27$

Which is multiple of \[27\].

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] ${{41}^{n}}-{{14}^{n}}$is true for $n=k$.

That is, \[P\text{ }\left( k \right):\] ${{41}^{k}}-{{14}^{k}}=27\lambda \text{ },where\text{ }\lambda \in \text{N    }.......\text{(i)}$

Now, we have to show that the given statement \[P\text{ }\left( n \right):\] ${{41}^{n}}-{{14}^{n}}$ is true for $n=k+1$.

\[P\text{ }\left( k+1 \right):\] ${{41}^{k+1}}-{{14}^{k+1}}$

${{41}^{k+1}}-{{14}^{k+1}}={{41}^{k}}\cdot 41-{{14}^{k}}\cdot 14$

$=(27\lambda +{{14}^{k}})\cdot 41-{{14}^{k}}\cdot 14 $ from equation (i)

$=27\lambda \cdot 41+{{14}^{k}}\cdot 41-{{14}^{k}}\cdot 14$

$=27\lambda \cdot 41+{{14}^{k}}(41-14)$

$=27\lambda \cdot 41+{{14}^{k}}\cdot 27$

$=27(41\lambda +{{14}^{k}})$

Hence, \[P\text{ }\left( k+1 \right):\] ${{41}^{k+1}}-{{14}^{k+1}}$ is multiple of \[27\].

Thus \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the given statement ${{41}^{n}}-{{14}^{n}}$ is multiple of \[27\].

11.  Using induction, prove that $\mathbf{{{10}^{n}}+3\cdot {{4}^{n+2}}+5\text{ }}$ is divisible by 9\forall n\in N .

Ans: In this Question we have to prove that ${{10}^{n}}+3\cdot {{4}^{n+2}}+5\text{ }$is divisible by \[9\] by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] ${{10}^{n}}+3\cdot {{4}^{n+2}}+5\text{ }$.

Checking if the statement is true or not for $n=1$.

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $10+3\cdot {{4}^{1+2}}+5=207$

Which is divisible by \[9\] .

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] ${{10}^{n}}+3\cdot {{4}^{n+2}}+5\text{ }$is true for $n=k$.

That is, $P\text{ }\left( k \right): {{10}^{k}}+3\cdot {{4}^{k+2}}+5\text{ }=9\lambda $,where $\lambda \in N$ .......(i)

Now, we have to show that the given statement \[P\text{ }\left( n \right):\] ${{10}^{n}}+3\cdot {{4}^{n+2}}+5\text{ }$ is true for$n=k+1$.

\[P\text{ }\left( k+1 \right):\] ${{10}^{k+1}}+3\cdot {{4}^{k+1+2}}+5\text{ }$

\[{{10}^{k+1}}+3\cdot {{4}^{k+3}}+5\text{ }\]

\[{{10}^{k}}\cdot 10+3\cdot {{4}^{k}}\cdot {{4}^{3}}+5\text{ }\]

$(9\lambda -3\cdot {{4}^{k+2}}-5)10+3\cdot {{4}^{k}}\cdot {{4}^{3}}+5\text{         }\!\!\{\!\!\text{ from equation (i) }\!\!\}\!\!\text{ }$

$90\lambda -30\cdot {{4}^{k+2}}-50+3\cdot {{4}^{k}}\cdot {{4}^{3}}+5\text{   }$

$90\lambda -30\cdot {{4}^{k+2}}-45+3\cdot {{4}^{k+2}}\cdot 4\text{  }$

$90\lambda -18\cdot {{4}^{k+2}}-45$

$9(10\lambda -2\cdot {{4}^{k+2}}-5)$

Hence, \[P\text{ }\left( k+1 \right):\] ${{10}^{k+1}}+3\cdot {{4}^{k+1+2}}+5\text{ }$ is divisible by \[9.\]

Thus, \[P\text{ }\left( k+1 \right)\] is true \[when\text{ }P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the given statement is true for every positive integer\[n\].

12. Prove that  $\mathbf{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{n}^{2}}=\dfrac{n(n+1)(2n+1)}{6}}$  .

Ans: In this Question we have to prove that ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{n}^{2}}=\dfrac{n(n+1)(2n+1)}{6}$ by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{n}^{2}}=\dfrac{n(n+1)(2n+1)}{6}$ .

Checking if the statement is true or not for $n=1$.

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $1=\dfrac{1(1+1)(2\cdot 1+1)}{6}$

$1=1$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{n}^{2}}=\dfrac{n(n+1)(2n+1)}{6}$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{k}^{2}}=\dfrac{k(k+1)(2k+1)}{6}$ ……….. (i)

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{n}^{2}}=\dfrac{n(n+1)(2n+1)}{6}$ Is true for $n=k+1$.

So, \[P\text{ }\left( k+1 \right):\] ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{k}^{2}}+{{(k+1)}^{2}}=\dfrac{(k+1)(k+2)(2k+3)}{6}$ .

Now, L.H.S $={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{k}^{2}}+{{(k+1)}^{2}}$

$=\dfrac{k(k+1)(2k+1)\text{ }}{6}\text{ }+{{(k+1)}^{2}}\text{  }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

$=\dfrac{k(k+1)(2k+1)+6{{(k+1)}^{2}}\text{ }}{6}\text{   }$

$=\dfrac{(k+1)[k(2k+1)+6(k+1)]\text{ }}{6}\text{   }$

$=\dfrac{(k+1)(2{{k}^{2}}\text{+k+6k+6) }}{6}\text{   }$

$=\dfrac{(k+1)(2{{k}^{2}}\text{+7k+6) }}{6}\text{   }$

$=\dfrac{(k+1)(k+2\text{)(2k+3)}}{6}\text{   }$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{k}^{2}}+{{(k+1)}^{2}}=\dfrac{(k+1)(k+2)(2k+3)}{6}$ is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the statement ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{n}^{2}}=\dfrac{n(n+1)(2n+1)}{6}$ is true.

13.  Prove that $\mathbf{1+3+{{3}^{2}}+...+{{3}^{n-1}}=\dfrac{{{3}^{n}}-1}{2}}$

Ans: In this Question we have to prove that $1+3+{{3}^{2}}+...+{{3}^{n-1}}=\dfrac{{{3}^{n}}-1}{2}$ by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $1+3+{{3}^{2}}+...+{{3}^{n-1}}=\dfrac{{{3}^{n}}-1}{2}$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $1=\dfrac{3-1}{2}$

$1=1$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $1+3+{{3}^{2}}+...+{{3}^{n-1}}=\dfrac{{{3}^{n}}-1}{2}$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] $1+3+{{3}^{2}}+...+{{3}^{k-1}}=\dfrac{{{3}^{k}}-1}{2}$ .........(i)

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $1+3+{{3}^{2}}+...+{{3}^{n-1}}=\dfrac{{{3}^{n}}-1}{2}$ Is true for $n=k+1$.

So, \[P\text{ }\left( k+1 \right):\] $1+3+{{3}^{2}}+...+{{3}^{k-1}}+{{3}^{k}}=\dfrac{{{3}^{k+1}}-1}{2}$ .

Now, L.H.S $=1+3+{{3}^{2}}+...+{{3}^{k-1}}+{{3}^{k}}$

$=\dfrac{{{3}^{k}}-1}{2}\text{+}{{\text{3}}^{k}}\text{  }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

$=\dfrac{{{3}^{k}}-1+2\times {{\text{3}}^{k}}}{2}\text{  }$

$=\dfrac{{{3}^{k}}(1+2)-1}{2}\text{  }$

$=\dfrac{{{3}^{k}}\times 3-1}{2}\text{  }$

$\dfrac{{{3}^{k+1}}-1}{2}\text{=R}\text{.H}\text{.S  }$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] $1+3+{{3}^{2}}+...+{{3}^{k-1}}+{{3}^{k}}=\dfrac{{{3}^{k+1}}-1}{2}$ is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the statement $1+3+{{3}^{2}}+...+{{3}^{n-1}}=\dfrac{{{3}^{n}}-1}{2}$ is true.

14. By induction, prove that \[\mathbf{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}>\dfrac{{{n}^{3}}}{2}\forall n\in N}\].

Ans: In this Question we have to prove that \[{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}>\dfrac{{{n}^{3}}}{3}\] by using the method of P.M.I.

Given, \[P\text{ }\left( n \right):\] \[{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}>\dfrac{{{n}^{3}}}{3}\] .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $1>\dfrac{1}{3}$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] \[{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}>\dfrac{{{n}^{3}}}{3}\] is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] \[{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{k}^{2}}>\dfrac{{{k}^{3}}}{3}\] ….. (i)

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] \[{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}>\dfrac{{{n}^{3}}}{3}\] Is true for $n=k+1$.

So, \[P\text{ }\left( k+1 \right):\] \[{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{k}^{2}}+{{(k+1)}^{2}}>\dfrac{{{(k+1)}^{3}}}{3}\]

Now, \[{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{k}^{2}}+{{(k+1)}^{2}}>\dfrac{{{(k+1)}^{3}}}{3}\]

$\dfrac{{{k}^{3}}}{3}+{{(k+1)}^{2}}>\dfrac{{{(k+1)}^{3}}}{3}$ from equation (1)

$\dfrac{{{k}^{3}}+3{{(k+1)}^{2}}}{3}>\dfrac{{{(k+1)}^{3}}}{3}\text{ }$

$\dfrac{{{k}^{3}}+3{{k}^{2}}+3+6k}{3}>\dfrac{{{(k+1)}^{3}}}{3}\text{ }$

\[\dfrac{{{(k+1)}^{3}}+(3k+2)}{3}>\dfrac{{{(k+1)}^{3}}}{3}\text{ }\]

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] \[{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{k}^{2}}+{{(k+1)}^{2}}>\dfrac{{{(k+1)}^{3}}}{3}\]  is true.

Thus \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the statement \[{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+.....+{{n}^{2}}>\dfrac{{{n}^{3}}}{3}\] is true for all  n $\in $ N .

15. Prove by PMI $\mathbf{{{(ab)}^{n}}={{a}^{n}}{{b}^{n}}}$ .

Ans: In this Question we have to prove that ${{(ab)}^{n}}={{a}^{n}}{{b}^{n}}$ by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] ${{(ab)}^{n}}={{a}^{n}}{{b}^{n}}$ .

Checking if the statement is true or not for $n=1$.

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $ab=ab$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] ${{(ab)}^{n}}={{a}^{n}}{{b}^{n}}$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] ${{(ab)}^{k}}={{a}^{k}}{{b}^{k}}$ ….. (i)

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] ${{(ab)}^{n}}={{a}^{n}}{{b}^{n}}$ Is true for $n=k+1$.

So, \[P\text{ }\left( k+1 \right):\] ${{(ab)}^{k+1}}={{a}^{k+1}}{{b}^{k+1}}$        

Now, L.H.S $={{(ab)}^{k+1}}$

$={{(ab)}^{k}}(ab)$

$={{a}^{k}}{{\text{b}}^{k}}\text{(ab)  }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

${{a}^{k+1}}{{\text{b}}^{k+1}}\text{=R}\text{.H}\text{.S  }$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] ${{(ab)}^{k+1}}={{a}^{k+1}}{{b}^{k+1}}$ is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the statement ${{(ab)}^{n}}={{a}^{n}}{{b}^{n}}$ is true.

16. Prove by PMI $\mathbf{a+ar+a{{r}^{2}}+......+a{{r}^{n-1}}=\dfrac{a({{r}^{n}}-1)}{r-1}}$

Ans: In this Question we have to prove that $a+ar+a{{r}^{2}}+......+a{{r}^{n-1}}=\dfrac{a({{r}^{n}}-1)}{r-1}$ by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $a+ar+a{{r}^{2}}+......+a{{r}^{n-1}}=\dfrac{a({{r}^{n}}-1)}{r-1}$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $a=a$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $a+ar+a{{r}^{2}}+......+a{{r}^{n-1}}=\dfrac{a({{r}^{n}}-1)}{r-1}$ is true for $n=k$.

That is, \[P\text{ }\left( k \right):\] $a+ar+a{{r}^{2}}+......+a{{r}^{k-1}}=\dfrac{a({{r}^{k}}-1)}{r-1}$  ……. (i)

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $a+ar+a{{r}^{2}}+......+a{{r}^{n-1}}=\dfrac{a({{r}^{n}}-1)}{r-1}$  Is true for $n=k+1$.

So, \[P\text{ }\left( k+1 \right):\] $a+ar+a{{r}^{2}}+......+a{{r}^{k-1}}+a{{r}^{k}}=\dfrac{a({{r}^{k+1}}-1)}{r-1}$        

Now, L.H.S $=a+ar+a{{r}^{2}}+......+a{{r}^{k-1}}+a{{r}^{k}}$

$=\dfrac{a({{r}^{k}}-1)}{r-1}+a{{r}^{k}}\text{ }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

$=\dfrac{a({{r}^{k}}-1)+a{{r}^{k+1}}-a{{r}^{k}}}{r-1}$

$=\dfrac{a{{r}^{k}}-a+a{{r}^{k+1}}-a{{r}^{k}}}{r-1}$

$=\dfrac{a{{r}^{k+1}}-a}{r-1}$

$\dfrac{a({{r}^{k+1}}-1)}{r-1}=R.H.S$

 Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] $a+ar+a{{r}^{2}}+......+a{{r}^{k-1}}+a{{r}^{k}}=\dfrac{a({{r}^{k+1}}-1)}{r-1}$  is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the statement $a+ar+a{{r}^{2}}+......+a{{r}^{n-1}}=\dfrac{a({{r}^{n}}-1)}{r-1}$ is true.

17. Prove that $\mathbf{{{x}^{2n}}-{{y}^{2n}}}$ is divisible by x+y.

Ans: In this Question we have to prove that ${{x}^{2n}}-{{y}^{2n}}$ is divisible by $x+y$ by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] ${{x}^{2n}}-{{y}^{2n}}$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] ${{x}^{2}}-{{y}^{2}}=(x+y)(x-y)$

Which is divisible by $x+y$ .

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] ${{x}^{2n}}-{{y}^{2n}}$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] \[{{x}^{2k}}-{{y}^{2k}}\text{ }=(x+y)\lambda\], where $\lambda \in N$ ….. (i)

Now, we have to show that the given statement \[P\text{ }\left( n \right):\] ${{x}^{2n}}-{{y}^{2n}}$ is true for $n=k+1$ .

\[P\text{ }\left( k+1 \right):\] ${{x}^{2(k+1)}}-{{y}^{2(k+1)}}$

$={{x}^{2k+2}}-{{y}^{2k+2}}$

$={{x}^{2k}}\cdot {{x}^{2}}-{{y}^{2k}}\cdot {{y}^{2}}$

\[=((x+y)\lambda +{{y}^{2k}}){{x}^{2}}-{{y}^{2k}}\cdot {{y}^{2}}\] from equation (i)

\[=\lambda (x+y){{x}^{2}}+{{x}^{2}}\cdot {{y}^{2k}}-{{y}^{2k}}\cdot {{y}^{2}}\]

\[=\lambda (x+y){{x}^{2}}+{{y}^{2k}}({{x}^{2}}-{{y}^{2}}\]

\[=\lambda (x+y){{x}^{2}}+{{y}^{2k}}(x-y\text{)(}x\text{+}y)\]

\[=(x+y)\{\lambda {{x}^{2}}+{{y}^{2k}}(x-y)\}\]

Hence, \[P\text{ }\left( k+1 \right):\] ${{x}^{2(k+1)}}-{{y}^{2(k+1)}}$ is divisible by $x+y$ .

Thus \[P\text{ }\left( k+1 \right)\] is true \[when\text{ }P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the given statement is true for every positive integer \[n\].

18. Prove that n(n+1)(2n+1) is divisible by 6.

Ans: In this Question we have to prove that $n(n+1)(2n+1)$ is divisible by \[\mathbf{6}\] by using the method of PMI.

Given, \[P\text{ }\left( n \right):\]$n(n+1)(2n+1)$ .

Checking if the statement is true or not for $n=1$.

So, For $n=1$

\[P\text{ }\left( 1 \right):\] \[1\cdot 2\cdot 3\]

Which is divisible by \[\mathbf{6}\].

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $n(n+1)(2n+1)$ is true for$n=k$.

That is, \[P\text{ }\left( k \right):\] \[k(k+1)(2k+1)\text{ }=6\lambda ,where\text{ }\lambda \in \text{N    }.......\text{(i)}\]

Now, we have to show that the given statement \[P\text{ }\left( n \right):\]$n(n+1)(2n+1)$ is true for$n=k+1$ .

\[P\text{ }\left( k+1 \right):\] \[(k+1)(k+2)(2k+3)\]

$=(k+1)(k+2)[(2k+1)+2]$

$=(k+1)(k+2)(2k+1)+2(k+1)(k+2)$

$=(k+2)[(k+1)(2k+1)]+2(k+1)(k+2)$

$=k(k+1)(2k+1)+2(k+1)(2k+1)+2(k+1)(k+2)$

\[=6\lambda +2(k+1)(2k+1)+2(k+1)(k+2)\] from equation (i)

\[=6\lambda +2(k+1)[2k+1+k+2]\text{  }\]

\[=6\lambda +2(k+1)[3k+3]\text{  }\]

\[=6\lambda +6(k+1)[k+1]\text{  }\]

\[=6[\lambda +{{(k+1)}^{2}}]\text{  }\]

Hence, \[P\text{ }\left( k+1 \right):\]\[(k+1)(k+2)(2k+3)\] is divisible by \[\mathbf{6}\] .

Thus \[P\text{ }\left( k+1 \right)\] Is true \[when\text{ }P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the given statement is true for every positive integer\[n\].

19. Show that $\mathbf{{{2}^{3n}}-1}$ is divisible by 7 .

Ans: In this Question we have to prove that ${{2}^{3n}}-1$ is divisible by $7$ by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] ${{2}^{3n}}-1$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] \[{{2}^{3}}-1=7\]

Which is divisible by $7$ .

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] ${{2}^{3n}}-1$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] \[{{2}^{3k}}-1\text{ }=7\lambda ,where\text{ }\lambda \in \text{N    }.......\text{(i)}\]

Now, we have to show that the given statement \[P\text{ }\left( n \right):\]${{2}^{3n}}-1$ is true for $n=k+1$ .

\[P\text{ }\left( k+1 \right):\] ${{2}^{3k+3}}-1$

$={{2}^{3k}}\cdot {{2}^{3}}-1$

$=(7\lambda +1)\cdot 8-1$ from equation (i)

$=56\lambda +8-1\text{   }$

$=56\lambda +7$

$=7(8\lambda +1)$

Hence,  \[P\text{ }\left( k+1 \right):\] ${{2}^{3k+3}}-1$ is divisible by $7$ .

Thus, \[P\text{ }\left( k+1 \right)\] is true \[when\text{ }P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the given statement is true for every positive integer \[n\].

20.  Prove by PMI $\mathbf{1\cdot 2\cdot 3+2\cdot 3\cdot 4+....+n\cdot (n+1)\cdot (n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}}$ .

Ans: In this Question we have to prove that  $1\cdot 2\cdot 3+2\cdot 3\cdot 4+....+n\cdot (n+1)\cdot (n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}$   by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $1\cdot 2\cdot 3+2\cdot 3\cdot 4+....+n\cdot (n+1)\cdot (n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $1\cdot 2\cdot 3=\dfrac{1(1+1)(1+2)(1+3)}{4}$

$6=6$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $1\cdot 2\cdot 3+2\cdot 3\cdot 4+....+n\cdot (n+1)\cdot (n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}$ is true for$n=k$ .

That is, \[P\text{ }\left( k \right):\] $1\cdot 2\cdot 3+2\cdot 3\cdot 4+....+k\cdot (k+1)\cdot (k+2)=\dfrac{k(k+1)(k+2)(k+3)}{4}\text{ }\ldots ..\left( \mathbf{i} \right)$

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $1\cdot 2\cdot 3+2\cdot 3\cdot 4+....+n\cdot (n+1)\cdot (n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}$ Is true for $n=k+1$.

So, \[P\text{ }\left( k+1 \right):\]\[1\cdot 2\cdot 3+2\cdot 3\cdot 4+....+k\cdot (k+1)\cdot (k+2)+(k+1)\cdot (k+2)\cdot (k+3)=\dfrac{(k+1)(k+2)(k+3)(k+4)}{4}\text{    }\]    

Now, L.H.S $=1\cdot 2\cdot 3+2\cdot 3\cdot 4+....+k\cdot (k+1)\cdot (k+2)+(k+1)\cdot (k+2)\cdot (k+3)$

$=\dfrac{k(k+1)(k+2)(k+3)}{4}\text{  }+(k+1)\cdot (k+2)\cdot (k+3)\text{ }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

$\dfrac{(k+1)(k+2)(k+3)(k+4)}{4}\text{ =R}\text{.H}\text{.S}$ 

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] \[1\cdot 2\cdot 3+2\cdot 3\cdot 4+....+k\cdot (k+1)\cdot (k+2)+(k+1)\cdot (k+2)\cdot (k+3)=\dfrac{(k+1)(k+2)(k+3)(k+4)}{4}\text{    }\] is true.

Thus \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore,

By P.M.I. the statement $1\cdot 2\cdot 3+2\cdot 3\cdot 4+....+n\cdot (n+1)\cdot (n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}$ is true.

21. Prove that $\mathbf{\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+...+\dfrac{1}{n\cdot (n+1)}=\dfrac{n}{n+1}}$ .

Ans: In this Question we have to prove that $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+...+\dfrac{1}{n\cdot (n+1)}=\dfrac{n}{n+1}$   by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+...+\dfrac{1}{n\cdot (n+1)}=\dfrac{n}{n+1}$

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $\dfrac{1}{2}=\dfrac{1}{2}$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+...+\dfrac{1}{n\cdot (n+1)}=\dfrac{n}{n+1}$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+...+\dfrac{1}{k\cdot (k+1)}=\dfrac{k}{k+1}\text{      }........\text{(i)}$

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+...+\dfrac{1}{n\cdot (n+1)}=\dfrac{n}{n+1}$ Is true for $n=k+1$ .

So, \[P\text{ }\left( k+1 \right):\]  $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+...+\dfrac{1}{k\cdot (k+1)}+\dfrac{1}{(k+1)\cdot (k+2)}=\dfrac{k+1}{k+2}$    

Now, L.H.S $=\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+...+\dfrac{1}{k\cdot (k+1)}+\dfrac{1}{(k+1)\cdot (k+2)}$

$=\dfrac{k}{k+1}\text{  }+\dfrac{1}{(k+1)\cdot (k+2)}\text{ }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

$=\dfrac{k(k+2)+1}{(k+1)\times (k+2)}\text{ }$

$=\dfrac{{{k}^{2}}+2k+1}{(k+1)\times (k+2)}\text{ }$

$=\dfrac{{{(k+1)}^{2}}}{(k+1)\cdot (k+2)}\text{ }$

$\dfrac{k+1}{k+2}\text{ =R}\text{.H}\text{.S}$ 

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+...+\dfrac{1}{k\cdot (k+1)}+\dfrac{1}{(k+1)\cdot (k+2)}=\dfrac{k+1}{k+2}$ is true.

Thus \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore, by P.M.I. the statement $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+...+\dfrac{1}{n\cdot (n+1)}=\dfrac{n}{n+1}$ is true.

22. Show that the sum of the first n odd natural number is $\mathbf{{{n}^{2}}}$ .

Ans: In this Question we have to prove that $1+3+5+.....+(2n-1)={{n}^{2}}$  by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $1+3+5+.....+(2n-1)={{n}^{2}}$

Checking if the statement is true or not for $n=1$.

So, For $n=1$

\[P\text{ }\left( 1 \right):\]$1=1$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $1+3+5+.....+(2n-1)={{n}^{2}}$ is true for $n=k$.

That is, \[P\text{ }\left( k \right):\] $1+3+5+.....+(2k-1)={{k}^{2}}$ ….. (i)

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $1+3+5+.....+(2n-1)={{n}^{2}}$ Is true for $n=k+1$ .

So, \[P\text{ }\left( k+1 \right):\] $1+3+5+.....+(2k+1)={{(k+1)}^{2}}$        

Now, L.H.S $=1+3+5+.....+(2k-1)+(2k+1)$

$={{k}^{2}}\text{+(2k+1)       }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

${{(k+1)}^{2}}\text{ =R}\text{.H}\text{.S}$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\]$1+3+5+.....+(2k+1)={{(k+1)}^{2}}$ is true.

Thus \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore, by P.M.I. the statement $1+3+5+.....+(2n-1)={{n}^{2}}$ is true.

23. Prove by PMI $\mathbf{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.......+{{n}^{3}}={{\left( \dfrac{n(n+1)}{2} \right)}^{2}}}$ .

Ans: In this Question we have to prove that ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.......+{{n}^{3}}={{\left( \dfrac{n(n+1)}{2} \right)}^{2}}$  by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.......+{{n}^{3}}={{\left( \dfrac{n(n+1)}{2} \right)}^{2}}$

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $1=1$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.......+{{n}^{3}}={{\left( \dfrac{n(n+1)}{2} \right)}^{2}}$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.......+{{k}^{3}}={{\left( \dfrac{k(k+1)}{2} \right)}^{2}}\text{    }..........\text{(i)}$

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.......+{{n}^{3}}={{\left( \dfrac{n(n+1)}{2} \right)}^{2}}$ Is true for $n=k+1$.

So, \[P\text{ }\left( k+1 \right):\] ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.......+{{k}^{3}}+{{(K+1)}^{3}}={{\left( \dfrac{(k+1)(k+2)}{2} \right)}^{2}}$    

Now, L.H.S $={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.......+{{k}^{3}}+{{(K+1)}^{3}}$

$={{\left( \dfrac{k(k+1)}{2} \right)}^{2}}+{{(K+1)}^{3}}\text{     }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

\[={{\dfrac{{{k}^{2}}(k+1)}{4}}^{2}}+{{(K+1)}^{3}}\]

\[=\dfrac{{{k}^{2}}{{(k+1)}^{2}}+4{{(K+1)}^{3}}}{4}\]

\[=\dfrac{{{(k+1)}^{2}}[{{k}^{2}}+4(K+1)]}{4}\]

\[=\dfrac{{{(k+1)}^{2}}[{{k}^{2}}+4K+4]}{4}\]

\[=\dfrac{{{(k+1)}^{2}}{{(k+2)}^{2}}}{4}\]

${{\left( \dfrac{(k+1)(k+2)}{2} \right)}^{2}}\text{ =R}\text{.H}\text{.S}$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\]${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.......+{{k}^{3}}+{{(K+1)}^{3}}={{\left( \dfrac{(k+1)(k+2)}{2} \right)}^{2}}$is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] Is true.

Therefore, by P.M.I. the statement ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.......+{{n}^{3}}={{\left( \dfrac{n(n+1)}{2} \right)}^{2}}$ is true.

24. Prove $\mathbf{\left( 1+\dfrac{3}{1} \right)\left( 1+\dfrac{5}{4} \right)\left( 1+\dfrac{7}{9} \right).......\left( 1+\dfrac{2n+1}{{{n}^{2}}} \right)={{(n+1)}^{2}}}$.

Ans: In this Question we have to prove that$\left( 1+\dfrac{3}{1} \right)\left( 1+\dfrac{5}{4} \right)\left( 1+\dfrac{7}{9} \right).......\left( 1+\dfrac{2n+1}{{{n}^{2}}} \right)={{(n+1)}^{2}}$  by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $\left( 1+\dfrac{3}{1} \right)\left( 1+\dfrac{5}{4} \right)\left( 1+\dfrac{7}{9} \right).......\left( 1+\dfrac{2n+1}{{{n}^{2}}} \right)={{(n+1)}^{2}}$

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $1=1$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $\left( 1+\dfrac{3}{1} \right)\left( 1+\dfrac{5}{4} \right)\left( 1+\dfrac{7}{9} \right).......\left( 1+\dfrac{2n+1}{{{n}^{2}}} \right)={{(n+1)}^{2}}$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] $\left( 1+\dfrac{3}{1} \right)\left( 1+\dfrac{5}{4} \right)\left( 1+\dfrac{7}{9} \right).......\left( 1+\dfrac{2k+1}{{{k}^{2}}} \right)={{(k+1)}^{2}}\text{   }..........\text{(i)}$

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $\left( 1+\dfrac{3}{1} \right)\left( 1+\dfrac{5}{4} \right)\left( 1+\dfrac{7}{9} \right).......\left( 1+\dfrac{2n+1}{{{n}^{2}}} \right)={{(n+1)}^{2}}$ Is true for $n=k+1$ .

So, \[P\text{ }\left( k+1 \right):\] $\left( 1+\dfrac{3}{1} \right)\left( 1+\dfrac{5}{4} \right)\left( 1+\dfrac{7}{9} \right).......\left( 1+\dfrac{2k+1}{{{k}^{2}}} \right)\left( 1+\dfrac{2k+3}{{{(k+1)}^{2}}} \right)={{(k+2)}^{2}}$    

Now, L.H.S $=\left( 1+\dfrac{3}{1} \right)\left( 1+\dfrac{5}{4} \right)\left( 1+\dfrac{7}{9} \right).......\left( 1+\dfrac{2k+1}{{{k}^{2}}} \right)\left( 1+\dfrac{2k+3}{{{(k+1)}^{2}}} \right)$

$={{(k+1)}^{2}}\left( 1+\dfrac{2k+3}{{{(k+1)}^{2}}} \right)$ from equation 1 

$={{(k+1)}^{2}}\left( \dfrac{{{(k+1)}^{2}}+2k+3}{{{(k+1)}^{2}}} \right)\text{ }$

$={{(k+1)}^{2}}\left( \dfrac{{{k}^{2}}+4k+4}{{{(k+1)}^{2}}} \right)\text{ }$

$={{k}^{2}}+4k+4$

${{(k+2)}^{2}}\text{=R}\text{.H}\text{.S}$ 

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] $\left( 1+\dfrac{3}{1} \right)\left( 1+\dfrac{5}{4} \right)\left( 1+\dfrac{7}{9} \right).......\left( 1+\dfrac{2k+1}{{{k}^{2}}} \right)\left( 1+\dfrac{2k+3}{{{(k+1)}^{2}}} \right)={{(k+2)}^{2}}$ is true.

Thus \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore, by P.M.I. the statement $\left( 1+\dfrac{3}{1} \right)\left( 1+\dfrac{5}{4} \right)\left( 1+\dfrac{7}{9} \right).......\left( 1+\dfrac{2n+1}{{{n}^{2}}} \right)={{(n+1)}^{2}}$ is true.

25. Prove that $\mathbf{{{3}^{2n+2}}-8n-9}$ is divisible by 8 .

Ans: In this Question we have to prove that ${{3}^{2n+2}}-8n-9$ is divisible by $8$ .by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] ${{3}^{2n+2}}-8n-9$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] \[{{3}^{4}}-8\cdot 1-9=81-17\]

$=64$

Which is divisible by $8$ .

Thus, \[P\text{ }\left( 1 \right)\]is true.

Let, \[P\text{ }\left( n \right):\]${{3}^{2n+2}}-8n-9$is true for$n=k$.

That is, \[P\text{ }\left( k \right):\]\[{{3}^{2k+2}}-8k-9\text{ }=8\lambda \], where $\lambda \in N$ ......(i)

Now, we have to show that the given statement \[P\text{ }\left( n \right):\]${{3}^{2n+2}}-8n-9$ is true for$n=k+1$ .

\[P\text{ }\left( k+1 \right):\] ${{3}^{2k+4}}-8(k+1)-9$

\[={{3}^{2k+4}}-8k-17\]

\[={{3}^{2k+2}}\cdot {{3}^{2}}-8k-17\]

$=(8\lambda +8k+9)\cdot 9-8k-17$ from equation (i)

$=72\lambda +72k+81-8k-17\text{ }$

$=72\lambda +64k+64\text{ }$

$=8(9\lambda +8k+8)\text{ }$

Hence, \[P\text{ }\left( k+1 \right):\] ${{3}^{2k+4}}-8(k+1)-9$ is divisible by $8$.

Thus, \[P\text{ }\left( k+1 \right)\] is true \[when\text{ }P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the given statement is true for every positive integer \[n\].

26. Prove by PMI $\mathbf{{{x}^{n}}-{{y}^{n}}}$ is divisible by $\mathbf{(x-y)\text{ }whenever\text{ x-y}\ne \text{0}}$ .

Ans: In this Question we have to prove that ${{x}^{n}}-{{y}^{n}}$ is divisible by $x-y$ by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] ${{x}^{n}}-{{y}^{n}}$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $x-y$

Which is divisible by $x-y$.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] ${{x}^{n}}-{{y}^{n}}$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] \[{{x}^{k}}-{{y}^{k}}\text{ }=(x-y)\lambda\], where $\lambda \in N$.......(i)

Now, we have to show that the given statement \[P\text{ }\left( n \right):\] ${{x}^{n}}-{{y}^{n}}$ is true for $n=k+1$ .

\[P\text{ }\left( k+1 \right):\] ${{x}^{(k+1)}}-{{y}^{(k+1)}}$

$={{x}^{k+1}}-{{y}^{k+1}}$

$={{x}^{k}}\cdot x-{{y}^{k}}\cdot y$

\[=((x-y)\lambda +{{y}^{k}})x-{{y}^{k}}\cdot y\] from equation (i) 

\[=\lambda (x-y)x+x\cdot {{y}^{k}}-{{y}^{k}}\cdot y\text{   }\]

\[=\lambda (x-y)x+{{y}^{k}}(x-y)\]

\[=(x-y)({{y}^{k}}+\lambda x)\]

Hence, \[P\text{ }\left( k+1 \right):\] ${{x}^{(k+1)}}-{{y}^{(k+1)}}$ is divisible by $x-y$

Thus, \[P\text{ }\left( k+1 \right)\] is true \[when\text{ }P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the given statement is true for every positive integer\[n\].

27. Prove $\mathbf{({{x}^{2n}}-1)}$ is divisible by (x-1).

Ans: In this Question we have to prove that $({{x}^{2n}}-1)$ is divisible by $(x-1)$ by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $({{x}^{2n}}-1)$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] ${{x}^{2}}-1=(x-1)(x+1)$

Which is divisible by $(x-1)$.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $({{x}^{2n}}-1)$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\]\[({{x}^{2k}}-1)\text{ }=(x-1)\lambda\], where $\lambda \in N$.......(i)

Now, we have to show that the given statement \[P\text{ }\left( n \right):\]$({{x}^{2n}}-1)$ is true for$n=k+1$.

\[P\text{ }\left( k+1 \right):\]$({{x}^{2k+2}}-1)$

$=({{x}^{2k}}\cdot {{x}^{2}}-1)$

\[=\{[(x-1)\lambda +1]\cdot {{x}^{2}}-1\} \] from equation (i)

\[=\{(x-1)\lambda \cdot {{x}^{2}}+{{x}^{2}}-1\}\text{  }\]

\[=(x-1)\lambda \cdot {{x}^{2}}+(x-1)(x+1)\text{  }\]

\[=(x-1)[\lambda \cdot {{x}^{2}}+(x+1)]\text{  }\]

Hence, \[P\text{ }\left( k+1 \right):\]$({{x}^{2k+2}}-1)$ is divisible by $(x-1)$

Thus, \[P\text{ }\left( k+1 \right)\] is true \[when\text{ }P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the given statement is true for every positive integer \[n\].

28.Prove $\mathbf{1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+......+\dfrac{1}{(1+2+....+n)}=+\dfrac{2n}{(n+1)}}$

Ans: In this Question we have to prove that $1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+......+\dfrac{1}{(1+2+....+n)}=+\dfrac{2n}{(n+1)}$  by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+......+\dfrac{1}{(1+2+....+n)}=+\dfrac{2n}{(n+1)}$

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $1=1$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+......+\dfrac{1}{(1+2+....+n)}=+\dfrac{2n}{(n+1)}$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] $1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+......+\dfrac{1}{(1+2+....+k)}=\dfrac{2k}{(k+1)}$ ………. (i)

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+......+\dfrac{1}{(1+2+....+n)}=+\dfrac{2n}{(n+1)}$ Is true for $n=k+1$ .

So, \[P\text{ }\left( k+1 \right):\] $1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+......+\dfrac{1}{(1+2+....+k)}+\dfrac{1}{(1+2+....+k+(k+1))}=\dfrac{2(k+1)}{(k+1)+1}$        

Now, L.H.S $=1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+......+\dfrac{1}{(1+2+....+k)}+\dfrac{1}{(1+2+....+k+(k+1))}$

$=\dfrac{2k}{(k+1)}+\dfrac{1}{(1+2+....+k+(k+1))}$ from equation (1)

 $\text{ }=\dfrac{2k}{(k+1)}+\dfrac{1}{\dfrac{(k+1)(k+2)}{2}}$ since the sum of n natural numbers = $\dfrac{n(n+1)}{2}$

$\text{ }=\dfrac{2k}{(k+1)}+\dfrac{2}{(k+1)(k+2)}\text{   }$

$\text{ }=\dfrac{2k}{(k+1)}+\dfrac{2}{(k+1)(k+2)}\text{   }$

$\text{ }=\dfrac{2k(k+2)+2}{(k+1)(k+2)}\text{   }$

$\text{ }=\dfrac{2{{k}^{2}}+4k+2}{(k+1)(k+2)}\text{   }$

$\text{ }=\dfrac{2({{k}^{2}}+2k+1)}{(k+1)(k+2)}\text{   }$

$\text{ }=\dfrac{2{{(k+1)}^{2}}}{(k+1)(k+2)}\text{   }$

$\dfrac{2(K+1)}{(k+2)}\text{=R}\text{.H}\text{.S}$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\]$1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+......+\dfrac{1}{(1+2+....+k)}+\dfrac{1}{(1+2+....+k+(k+1))}=\dfrac{2(k+1)}{(k+1)+1}$is true.

Thus \[P\text{ }\left( k+1 \right)\]is true when \[P\text{ }\left( k \right)\]is true.

Therefore, by P.M.I. the statement $1+\dfrac{1}{(1+2)}+\dfrac{1}{(1+2+3)}+......+\dfrac{1}{(1+2+....+n)}=+\dfrac{2n}{(n+1)}$ is true.

29. Prove $\mathbf{1\cdot 3+3\cdot 5+5\cdot 7+.....+(2n-1)(2n+1)=\dfrac{n(4{{n}^{2}}+6n-1)}{3}}$ .

Ans: In this Question we have to prove that $1\cdot 3+3\cdot 5+5\cdot 7+.....+(2n-1)(2n+1)=\dfrac{n(4{{n}^{2}}+6n-1)}{3}$  by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $1\cdot 3+3\cdot 5+5\cdot 7+.....+(2n-1)(2n+1)=\dfrac{n(4{{n}^{2}}+6n-1)}{3}$

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $3=\dfrac{1(4\cdot {{1}^{2}}+6\cdot 1-1)}{3}$

$3=\dfrac{9}{3}$

$3=3$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $1\cdot 3+3\cdot 5+5\cdot 7+.....+(2n-1)(2n+1)=\dfrac{n(4{{n}^{2}}+6n-1)}{3}$  is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] $1\cdot 3+3\cdot 5+5\cdot 7+.....+(2k-1)(2k+1)=\dfrac{k(4{{k}^{2}}+6k-1)}{3}$

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $1\cdot 3+3\cdot 5+5\cdot 7+.....+(2n-1)(2n+1)=\dfrac{n(4{{n}^{2}}+6n-1)}{3}$ Is true for $n=k+1$ .

So, \[P\text{ }\left( k+1 \right):\] $1\cdot 3+3\cdot 5+5\cdot 7+.....+(2(k+1)-1)(2(k+1)+1)=\dfrac{(k+1)[4{{(k+1)}^{2}}+6(k+1)-1]}{3}$        

Now, L.H.S $=1\cdot 3+3\cdot 5+5\cdot 7+.....+(2k-1)(2k+1)+(2(k+1)-1)(2(k+1)+1)$

$=\dfrac{k(4{{k}^{2}}+6k-1)}{3}+(2(k+1)-1)(2(k+1)+1)$ from equation (1)

$=\dfrac{k(4{{k}^{2}}+6k-1)}{3}+(2k+1)(2k+3)\text{ }$ 

$=\dfrac{k(4{{k}^{2}}+6k-1)+3(2k+1)(2k+3)}{3}\text{ }$

$=\dfrac{4{{k}^{3}}+18{{k}^{2}}+23k+9}{3}\text{ }$

 $\dfrac{(k+1)(4{{k}^{2}}+14k+9)}{3}\text{ =R}\text{.H}\text{.S}$

 Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] $1\cdot 3+3\cdot 5+5\cdot 7+.....+(2(k+1)-1)(2(k+1)+1)=\dfrac{(k+1)[4{{(k+1)}^{2}}+6(k+1)-1]}{3}$ is true.

Thus \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore, by P.M.I. the statement $1\cdot 3+3\cdot 5+5\cdot 7+.....+(2n-1)(2n+1)=\dfrac{n(4{{n}^{2}}+6n-1)}{3}$ is true.

30. Prove by PMI $\mathbf{3\cdot {{2}^{2}}+{{3}^{2}}\cdot {{2}^{3}}+{{3}^{3}}\cdot {{2}^{4}}+....+{{3}^{n}}\cdot {{2}^{n+1}}=\dfrac{12}{5}({{6}^{n}}-1)}$

Ans: In this Question we have to prove that $3\cdot {{2}^{2}}+{{3}^{2}}\cdot {{2}^{3}}+{{3}^{3}}\cdot {{2}^{4}}+....+{{3}^{n}}\cdot {{2}^{n+1}}=\dfrac{12}{5}({{6}^{n}}-1)$  by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $3\cdot {{2}^{2}}+{{3}^{2}}\cdot {{2}^{3}}+{{3}^{3}}\cdot {{2}^{4}}+....+{{3}^{n}}\cdot {{2}^{n+1}}=\dfrac{12}{5}({{6}^{n}}-1)$

Checking if the statement is true or not for $n=1$ .

So, For$n=1$

\[P\text{ }\left( 1 \right):\]$3\cdot 4=\dfrac{12}{5}(6-1)$

$12=12$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $3\cdot {{2}^{2}}+{{3}^{2}}\cdot {{2}^{3}}+{{3}^{3}}\cdot {{2}^{4}}+....+{{3}^{n}}\cdot {{2}^{n+1}}=\dfrac{12}{5}({{6}^{n}}-1)$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] $3\cdot {{2}^{2}}+{{3}^{2}}\cdot {{2}^{3}}+{{3}^{3}}\cdot {{2}^{4}}+....+{{3}^{k}}\cdot {{2}^{k+1}}=\dfrac{12}{5}({{6}^{k}}-1)\text{      }........\text{(i)}$

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $3\cdot {{2}^{2}}+{{3}^{2}}\cdot {{2}^{3}}+{{3}^{3}}\cdot {{2}^{4}}+....+{{3}^{n}}\cdot {{2}^{n+1}}=\dfrac{12}{5}({{6}^{n}}-1)$ Is true for $n=k+1$ .

So, \[P\text{ }\left( k+1 \right):\] $3\cdot {{2}^{2}}+{{3}^{2}}\cdot {{2}^{3}}+{{3}^{3}}\cdot {{2}^{4}}+....+{{3}^{k}}\cdot {{2}^{k+1}}+{{3}^{k+1}}\cdot {{2}^{k+2}}=\dfrac{12}{5}({{6}^{k+1}}-1)$        

Now, L.H.S $=3\cdot {{2}^{2}}+{{3}^{2}}\cdot {{2}^{3}}+{{3}^{3}}\cdot {{2}^{4}}+....+{{3}^{k}}\cdot {{2}^{k+1}}+{{3}^{k+1}}\cdot {{2}^{k+2}}$

$=\dfrac{12}{5}({{6}^{k}}-1)+{{3}^{k+1}}\cdot {{2}^{k+2}}$  from equation (1)

$=\dfrac{2}{5}\cdot {{6}^{k}}\cdot 6-\dfrac{12}{5}+{{3}^{k+1}}\cdot {{2}^{k+1}}\cdot 2$ 

$=\dfrac{2}{5}\cdot {{6}^{k+1}}-\dfrac{12}{5}+{{6}^{k+1}}\cdot 2$

$={{6}^{k+1}}\left( \dfrac{2}{5}+2 \right)-\dfrac{12}{5}$

$\dfrac{12}{5}({{6}^{k+1}}-\text{1) =R}\text{.H}\text{.S}$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] $3\cdot {{2}^{2}}+{{3}^{2}}\cdot {{2}^{3}}+{{3}^{3}}\cdot {{2}^{4}}+....+{{3}^{k}}\cdot {{2}^{k+1}}+{{3}^{k+1}}\cdot {{2}^{k+2}}=\dfrac{12}{5}({{6}^{k+1}}-1)$ is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore, by P.M.I. the statement $3\cdot {{2}^{2}}+{{3}^{2}}\cdot {{2}^{3}}+{{3}^{3}}\cdot {{2}^{4}}+....+{{3}^{n}}\cdot {{2}^{n+1}}=\dfrac{12}{5}({{6}^{n}}-1)$ is true.

31. Prove  $\mathbf{1\cdot 3+2\cdot {{3}^{2}}+3\cdot {{3}^{3}}+....+n\cdot {{3}^{n}}=\dfrac{(2n-1){{3}^{n+1}}+3}{4}}$  .

Ans: In this Question we have to prove that $1\cdot 3+2\cdot {{3}^{2}}+3\cdot {{3}^{3}}+....+n\cdot {{3}^{n}}=\dfrac{(2n-1){{3}^{n+1}}+3}{4}$   by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $1\cdot 3+2\cdot {{3}^{2}}+3\cdot {{3}^{3}}+....+n\cdot {{3}^{n}}=\dfrac{(2n-1){{3}^{n+1}}+3}{4}$

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $3=\dfrac{(2-1){{3}^{2}}+3}{2}$

$3=3$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $1\cdot 3+2\cdot {{3}^{2}}+3\cdot {{3}^{3}}+....+n\cdot {{3}^{n}}=\dfrac{(2n-1){{3}^{n+1}}+3}{4}$ is true for $n=k$.

That is, \[P\text{ }\left( k \right):\] $1\cdot 3+2\cdot {{3}^{2}}+3\cdot {{3}^{3}}+....+k\cdot {{3}^{k}}=\dfrac{(2k-1){{3}^{k+1}}+3}{4}\text{ }............\text{(i)}$

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $1\cdot 3+2\cdot {{3}^{2}}+3\cdot {{3}^{3}}+....+n\cdot {{3}^{n}}=\dfrac{(2n-1){{3}^{n+1}}+3}{4}$ Is true for $n=k+1$.

So, \[P\text{ }\left( k+1 \right):\] $1\cdot 3+2\cdot {{3}^{2}}+3\cdot {{3}^{3}}+....+k\cdot {{3}^{k}}+(k+1)\cdot {{3}^{k+1}}=\dfrac{(2k+1){{3}^{k+2}}+3}{4}\text{ }$    

Now, L.H.S $=1\cdot 3+2\cdot {{3}^{2}}+3\cdot {{3}^{3}}+....+k\cdot {{3}^{k}}+(k+1)\cdot {{3}^{k+1}}$

$=\dfrac{(2k-1){{3}^{k+1}}+3}{4}+(k+1)\cdot {{3}^{k+1}}\text{ }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

$=\dfrac{(2k-1){{3}^{k+1}}+3+4(k+1){{3}^{k+1}}}{4}$ 

$=\dfrac{(2k-1+4k+4){{3}^{k+1}}+3}{4}$

$=\dfrac{(6k+3){{3}^{k+1}}+3}{4}$

$\dfrac{(2k+1){{3}^{k+2}}+3}{4}\text{ =R}\text{.H}\text{.S}$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] $1\cdot 3+2\cdot {{3}^{2}}+3\cdot {{3}^{3}}+....+k\cdot {{3}^{k}}+(k+1)\cdot {{3}^{k+1}}=\dfrac{(2k+1){{3}^{k+2}}+3}{4}\text{ }$is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore, by P.M.I. the statement $1\cdot 3+2\cdot {{3}^{2}}+3\cdot {{3}^{3}}+....+n\cdot {{3}^{n}}=\dfrac{(2n-1){{3}^{n+1}}+3}{4}$ is true.

32. Prove $\mathbf{\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\dfrac{1}{7\cdot 9}+.......+\dfrac{1}{(2n+1)(2n+3)}=\dfrac{n}{3(2n+3)}}$ .

Ans: In this Question we have to prove that $\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\dfrac{1}{7\cdot 9}+.......+\dfrac{1}{(2n+1)(2n+3)}=\dfrac{n}{3(2n+3)}$  by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\dfrac{1}{7\cdot 9}+.......+\dfrac{1}{(2n+1)(2n+3)}=\dfrac{n}{3(2n+3)}$

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $\dfrac{1}{15}=\dfrac{1}{15}$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\dfrac{1}{7\cdot 9}+.......+\dfrac{1}{(2n+1)(2n+3)}=\dfrac{n}{3(2n+3)}$ is true for $n=k$.

That is, \[P\text{ }\left( k \right):\]  $\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\dfrac{1}{7\cdot 9}+.......+\dfrac{1}{(2k+1)(2k+3)}=\dfrac{k}{3(2k+3)}\text{ }............\text{(i)}$

Now, we have to show that the given statement.

\[P\text{ }\left( n \right):\] $\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\dfrac{1}{7\cdot 9}+.......+\dfrac{1}{(2n+1)(2n+3)}=\dfrac{n}{3(2n+3)}$ Is true for $n=k+1$ .

So, \[P\text{ }\left( k+1 \right):\]$\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\dfrac{1}{7\cdot 9}+.......+\dfrac{1}{(2k+3)(2k+5)}=\dfrac{k+1}{3(2k+5)}\text{ }$

Now, L.H.S $\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\dfrac{1}{7\cdot 9}+.......+\dfrac{1}{(2k+3)(2k+5)}+\dfrac{1}{(2k+3)(2k+5)}$

$=\dfrac{k}{3(2k+3)}+\dfrac{1}{(2k+3)(2k+5)}\text{ }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

$=\dfrac{k(2k+5)+3}{3(2k+3)(2k+5)}\text{ }$ 

$=\dfrac{2{{k}^{2}}+5k+3}{3(2k+3)(2k+5)}\text{ }$

$=\dfrac{(2k+3)(k+1)}{3(2k+3)(2k+5)}\text{ }$

$\dfrac{(k+1)}{3(2k+5)}\text{  =R}\text{.H}\text{.S}$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] $\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\dfrac{1}{7\cdot 9}+.......+\dfrac{1}{(2k+3)(2k+5)}=\dfrac{k+1}{3(2k+5)}\text{ }$is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore, by P.M.I. the statement $\dfrac{1}{3\cdot 5}+\dfrac{1}{5\cdot 7}+\dfrac{1}{7\cdot 9}+.......+\dfrac{1}{(2n+1)(2n+3)}=\dfrac{n}{3(2n+3)}$ is true.

33. The sum of the cubes of three consecutive natural number is divisible by 9 .

Ans: In this Question we have to prove that $[{{n}^{3}}+{{(n+1)}^{3}}+{{(n+2)}^{3}}]$ is divisible by \[9\] by using the method of PMI.

Given, \[P\text{ }\left( n \right):\]$[{{n}^{3}}+{{(n+1)}^{3}}+{{(n+2)}^{3}}]$.

Checking if the statement is true or not for$n=1$.

So, For $n=1$

\[P\text{ }\left( 1 \right):\] ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}=243$

Which is divisible by \[9\] .

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\]$[{{n}^{3}}+{{(n+1)}^{3}}+{{(n+2)}^{3}}]$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] \[[{{k}^{3}}+{{(k+1)}^{3}}+{{(k+2)}^{3}}]=9\lambda ,where\text{ }\lambda \in \text{N    }.......\text{(i)}\]

Now, we have to show that the given statement \[P\text{ }\left( n \right):\]$[{{n}^{3}}+{{(n+1)}^{3}}+{{(n+2)}^{3}}]$ is true for $n=k+1$ .

\[P\text{ }\left( k+1 \right):\] $[{{(k+1)}^{3}}+{{(k+2)}^{3}}+{{(k+3)}^{3}}]$

$=[{{(k+1)}^{3}}+{{(k+2)}^{3}}+{{k}^{3}}+9{{k}^{2}}+27k+27]$

\[=[9\lambda +9{{k}^{2}}+27k+27\] from equation (i) 

\[=9[\lambda +{{k}^{2}}+3k+3]\text{  }\]

Which is divisible by \[9\] .

Hence, \[P\text{ }\left( k+1 \right):\] $[{{(k+1)}^{3}}+{{(k+2)}^{3}}+{{(k+3)}^{3}}]$ is divisible by \[9\] .

Thus, \[P\text{ }\left( k+1 \right)\] is true \[when\text{ }P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the given statement is true for every positive integer \[n\].

34. Prove that $\mathbf{{{12}^{n}}+{{25}^{n-1}}}$ is divisible by 13 .

Ans: In this Question we have to prove that ${{12}^{n}}+{{25}^{n-1}}$ is divisible by $13$ by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] ${{12}^{n}}+{{25}^{n-1}}$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $12+{{(25)}^{0}}=13$

Which is divisible by $13$ .

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] ${{12}^{n}}+{{25}^{n-1}}$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] \[{{12}^{k}}+{{25}^{k-1}}=13\lambda ,where\text{ }\lambda \in \text{N    }.......\text{(i)}\]

Now, we have to show that the given statement \[P\text{ }\left( n \right):\] ${{12}^{n}}+{{25}^{n-1}}$ is true for $n=k+1$.

\[P\text{ }\left( k+1 \right):\]${{12}^{k+1}}+{{25}^{k}}$

$={{12}^{k}}\cdot 12+{{25}^{k}}$

\[=(13\lambda -{{25}^{k-1}})12+{{25}^{k}}\] from equation (i)

\[=13\cdot 12\lambda -12\cdot {{25}^{k-1}}+{{25}^{k}}\text{ }\]

\[=13\cdot 12\lambda +{{25}^{k-1}}\text{(}-12+25\text{) }\]

\[=13\cdot 12\lambda +13\cdot {{25}^{k-1}}\]

\[=13(12\lambda +{{25}^{k-1}})\]

Which is divisible by $13$ .

Hence, \[P\text{ }\left( k+1 \right):\] ${{12}^{k+1}}+{{25}^{k}}$ is divisible by $13$ .

Thus \[P\text{ }\left( k+1 \right)\] is true \[when\text{ }P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the given statement is true for every positive integer \[n\].

35. Prove $\mathbf{{{11}^{n+2}}+{{12}^{2n+1}}}$ is divisible by 133 .

Ans: In this Question we have to prove that ${{11}^{n+2}}+{{12}^{2n+1}}$ is divisible by $133$ by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] ${{11}^{n+2}}+{{12}^{2n+1}}$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\]${{11}^{3}}+{{12}^{3}}=1331+1728$

$=3059$

Which is divisible by$133$.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] ${{11}^{n+2}}+{{12}^{2n+1}}$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] \[{{11}^{k+2}}+{{12}^{2k+1}}=133\lambda ,where\text{ }\lambda \in \text{N    }.......\text{(i)}\]

Now, we have to show that the given statement \[P\text{ }\left( n \right):\] ${{11}^{n+2}}+{{12}^{2n+1}}$ is true for $n=k+1$ .

\[P\text{ }\left( k+1 \right):\] ${{11}^{k+3}}+{{12}^{2k+3}}$

\[={{11}^{k+2}}\times 11+{{12}^{2k+1}}\times {{12}^{2}}\]

\[=(133\lambda -{{12}^{2k+1}})11+{{12}^{2k+1}}\times {{12}^{2}}\] from equation (i) 

\[=133\times 11\lambda -11\times {{12}^{2k+1}}+{{12}^{2k+1}}\times {{12}^{2}}\text{  }\]

\[=133\cdot 11\lambda -{{12}^{2k+1}}(-11+{{12}^{2}}\text{)  }\]

\[=133\cdot 11\lambda -{{12}^{2k+1}}\cdot 133\]

\[=133\cdot (11\lambda -{{12}^{2k+1}})\]

Which is divisible by $133$ .

Hence, \[P\text{ }\left( k+1 \right):\] ${{11}^{k+3}}+{{12}^{2k+3}}$ is divisible by $133$ .

Thus, \[P\text{ }\left( k+1 \right)\] is true \[when\text{ }P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the given statement is true for every positive integer \[n\].

36. Prove $\mathbf{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}=\dfrac{{{n}^{2}}{{(n+1)}^{2}}}{4}}$ .

Ans: In this Question we have to prove that ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}=\dfrac{{{n}^{2}}{{(n+1)}^{2}}}{4}$  by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}=\dfrac{{{n}^{2}}{{(n+1)}^{2}}}{4}$

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $1=\dfrac{1{{(1+1)}^{2}}}{4}$

$1=1$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}=\dfrac{{{n}^{2}}{{(n+1)}^{2}}}{4}$ is true for $n=k$.

That is, \[P\text{ }\left( k \right):\] ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{k}^{3}}=\dfrac{{{k}^{2}}{{(k+1)}^{2}}}{4}\text{ }............\text{(i)}$

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}=\dfrac{{{n}^{2}}{{(n+1)}^{2}}}{4}$ Is true for $n=k+1$ .

So, \[P\text{ }\left( k+1 \right):\] ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{k}^{3}}+{{(k+1)}^{3}}=\dfrac{{{(k+1)}^{2}}{{(k+2)}^{2}}}{4}$        

Now, L.H.S ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{k}^{3}}+{{(k+1)}^{3}}$

$=\dfrac{{{k}^{2}}{{(k+1)}^{2}}}{4}+{{(k+1)}^{3}}\text{ }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

$=\dfrac{{{k}^{2}}{{(k+1)}^{2}}+4{{(k+1)}^{3}}}{4}$ 

$=\dfrac{{{(k+1)}^{2}}[{{k}^{2}}+4(k+1)]}{4}$

$=\dfrac{{{(k+1)}^{2}}[{{k}^{2}}+4k+4]}{4}$

$\dfrac{{{(k+1)}^{2}}{{(k+2)}^{2}}}{4}\text{ =R}\text{.H}\text{.S}$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\]${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{k}^{3}}+{{(k+1)}^{3}}=\dfrac{{{(k+1)}^{2}}{{(k+2)}^{2}}}{4}$is true.

Thus \[P\text{ }\left( k+1 \right)\]is true when \[P\text{ }\left( k \right)\]is true.

Therefore, by P.M.I. the statement ${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+......+{{n}^{3}}=\dfrac{{{n}^{2}}{{(n+1)}^{2}}}{4}$ is true.

37. Prove $\mathbf{(a)+(a+d)+(a+2d)+......+[a+(n-1)d]=\dfrac{n}{2}[2a+(n-1)d]}$ .

Ans: In this Question we have to prove that $(a)+(a+d)+(a+2d)+......+[a+(n-1)d]=\dfrac{n}{2}[2a+(n-1)d]$  by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $(a)+(a+d)+(a+2d)+......+[a+(n-1)d]=\dfrac{n}{2}[2a+(n-1)d]$

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $a=a$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $(a)+(a+d)+(a+2d)+......+[a+(n-1)d]=\dfrac{n}{2}[2a+(n-1)d]$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] $(a)+(a+d)+(a+2d)+......+[a+(k-1)d]=\dfrac{k}{2}[2a+(k-1)d]............\text{(i)}$

Now, we have to show that the given statement.

\[P\text{ }\left( n \right):\] $(a)+(a+d)+(a+2d)+......+[a+(n-1)d]=\dfrac{n}{2}[2a+(n-1)d]$ Is true for $n=k+1$.

So, \[P\text{ }\left( k+1 \right):\] $(a)+(a+d)+(a+2d)+......+[a+(k-1)d]+[a+k\cdot d]=\dfrac{k+1}{2}[2a+k\cdot d]$    

Now, L.H.S $(a)+(a+d)+(a+2d)+......+[a+(k-1)d]+[a+k\cdot d]$

$=\dfrac{k}{2}[2a+(k-1)d]+[a+k\cdot d]\text{ }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

$=ka+\dfrac{k}{2}(k-1)d+[a+k\cdot d]$ 

$=ka+\dfrac{d{{k}^{2}}-dk}{2}+a+k\cdot d$

$=\dfrac{2ka+d{{k}^{2}}-dk+2a+2k\cdot d}{2}$

$=\dfrac{2ka+d{{k}^{2}}+dk+2a}{2}$

$=\dfrac{2a(k+1)+dk(k+1)}{2}$

$\dfrac{(k+1)(2a+dk)}{2}\text{=R}\text{.H}\text{.S}$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] $(a)+(a+d)+(a+2d)+......+[a+(k-1)d]+[a+k\cdot d]=\dfrac{k+1}{2}[2a+k\cdot d]$ is true.

Thus \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore, by P.M.I. the statement $(a)+(a+d)+(a+2d)+......+[a+(n-1)d]=\dfrac{n}{2}[2a+(n-1)d]$ is true.

38. Prove that $\mathbf{{{2}^{n}}>n\ }$for all  positive integers n.

Ans: In this Question we have to prove that ${{2}^{n}}>n$ by using the method of P.M.I.

Given, \[P\text{ }\left( n \right):\] ${{2}^{n}}>n$ .

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\] $2>1$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\]${{2}^{n}}>n$ is true for$n=k$.

That is, \[P\text{ }\left( k \right):\]\[{{2}^{k}}>k\text{       }.......\text{(i)}\]

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] ${{2}^{n}}>n$ Is true for $n=k+1$ .

So, \[P\text{ }\left( k+1 \right):\]${{2}^{k+1}}>k+1$

Now, ${{2}^{k+1}}>k+1$

${{2}^{k}}\times 2>2k$ from equation (1)

\[{{2}^{k}}\times 2>k+k>k+1\]

So, ${{2}^{k+1}}>k+1$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] ${{2}^{k+1}}>k+1$ is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore by P.M.I. the statement ${{2}^{n}}>n$ is true for all n $\in $ N.

39. Prove $\mathbf{\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+....+\dfrac{1}{n\cdot (n-1)}=\dfrac{n}{n+1}}$ .

Ans: In this Question we have to prove that $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+....+\dfrac{1}{n\cdot (n+1)}=\dfrac{n}{n+1}$  by using the method of PMI.

Given, \[P\text{ }\left( n \right):\] $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+....+\dfrac{1}{n\cdot (n+1)}=\dfrac{n}{n+1}$

Checking if the statement is true or not for $n=1$ .

So, For $n=1$

\[P\text{ }\left( 1 \right):\]$\dfrac{1}{2}=\dfrac{1}{2}$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+....+\dfrac{1}{n\cdot (n+1)}=\dfrac{n}{n+1}$ is true for $n=k$ .

That is, \[P\text{ }\left( k \right):\] $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+....+\dfrac{1}{k\cdot (k+1)}=\dfrac{k}{k+1}...........\text{(i)}$

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+....+\dfrac{1}{n\cdot (n+1)}=\dfrac{n}{n+1}$ Is true for $n=k+1$ .

So, \[P\text{ }\left( k+1 \right):\]$\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+....+\dfrac{1}{k(k+1)}+\dfrac{1}{(k+2)\cdot (k)}=\dfrac{k+1}{k+2}$

Now, L.H.S $=\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+....+\dfrac{1}{k(k+1)}+\dfrac{1}{(k+2)\cdot (k+1)}$

$=\dfrac{k}{k+1}+\dfrac{1}{(k+1)\cdot (k+2)}\text{ }........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

$=\dfrac{k(k+2)+1}{(k+1)\cdot (k+2)}\text{ }$ 

$=\dfrac{{{k}^{2}}+2k+1}{(k+1)\cdot (k+2)}\text{ }$

$\dfrac{(k+1)}{(k+2)}\text{=R}\text{.H}\text{.S }$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+....+\dfrac{1}{k(k+1)}+\dfrac{1}{(k+2)\cdot (k)}=\dfrac{k+1}{k+2}$ is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore, by P.M.I. the statement $\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+....+\dfrac{1}{n\cdot (n+1)}=\dfrac{n}{n+1}$ is true.

40. Prove $\mathbf{\dfrac{1}{3\cdot 6}+\dfrac{1}{6\cdot 9}+\dfrac{1}{9\cdot 12}+.........+\dfrac{1}{3n\cdot 3(n+1)}=\dfrac{n}{9\cdot (n+1)}}$ .

Ans: In this Question we have to prove that $\dfrac{1}{3\cdot 6}+\dfrac{1}{6\cdot 9}+\dfrac{1}{9\cdot 12}+.........+\dfrac{1}{3n\cdot 3(n+1)}=\dfrac{n}{9\cdot (n+1)}$  by using the method of PMI.

Given, \[P\text{ }\left( n \right):\]$\dfrac{1}{3\cdot 6}+\dfrac{1}{6\cdot 9}+\dfrac{1}{9\cdot 12}+.........+\dfrac{1}{3n\cdot 3(n+1)}=\dfrac{n}{9\cdot (n+1)}$

Checking if the statement is true or not for $n=1$.

So, For$n=1$

\[P\text{ }\left( 1 \right):\] $\dfrac{1}{18}=\dfrac{1}{18}$

Which is true.

Thus, \[P\text{ }\left( 1 \right)\] is true.

Let, \[P\text{ }\left( n \right):\] $\dfrac{1}{3\times 6}+\dfrac{1}{6\times 9}+\dfrac{1}{9\times 12}+.........+\dfrac{1}{3n\times 3(n+1)}=\dfrac{n}{9\times (n+1)}$ is true for$n=k$.

That is, \[P\text{ }\left( k \right):\] $\dfrac{1}{3\times 6}+\dfrac{1}{6\times 9}+\dfrac{1}{9\times 12}+.........+\dfrac{1}{3k\times 3(k+1)}=\dfrac{k}{9\times (k+1)}...........\text{(i)}$

Now, we have to show that the given statement

\[P\text{ }\left( n \right):\] $\dfrac{1}{3\times 6}+\dfrac{1}{6\times 9}+\dfrac{1}{9\times 12}+.........+\dfrac{1}{3n\times 3(n+1)}=\dfrac{n}{9\times (n+1)}$ Is true for $n=k+1$ .

So, \[P\text{ }\left( k+1 \right):\] $\dfrac{1}{3\cdot 6}+\dfrac{1}{6\cdot 9}+\dfrac{1}{9\cdot 12}+.........+\dfrac{1}{3(k+1)\cdot 3(k+2)}=\dfrac{k+1}{9\cdot (k+2)}$    

Now, L.H.S $=\dfrac{1}{3\cdot 6}+\dfrac{1}{6\cdot 9}+\dfrac{1}{9\cdot 12}+.........+\dfrac{1}{3(k)\cdot 3(k+1)}+\dfrac{1}{3(k+1)\cdot 3(k+2)}$

$=\dfrac{k}{9\cdot (k+1)}+\dfrac{1}{3(k+1)\cdot 3(k+2)}........\text{ }\!\!\{\!\!\text{ from equation 1 }\!\!\}\!\!\text{ }$

$=\dfrac{k(k+2)+1}{9(k+1)(k+2)}$ 

$=\dfrac{{{k}^{2}}+2k+1}{9(k+1)\cdot (k+2)}\text{ }$

$=\dfrac{{{(k+1)}^{2}}}{9(k+1)\cdot (k+2)}\text{ }$

$\dfrac{(k+1)}{9(k+2)}\text{=R}\text{.H}\text{.S }$

Which is true.

Hence, \[P\text{ }\left( k+1 \right):\] $\dfrac{1}{3\cdot 6}+\dfrac{1}{6\cdot 9}+\dfrac{1}{9\cdot 12}+.........+\dfrac{1}{3(k+1)\cdot 3(k+2)}=\dfrac{k+1}{9\cdot (k+2)}$ is true.

Thus, \[P\text{ }\left( k+1 \right)\] is true when \[P\text{ }\left( k \right)\] is true.

Therefore, by P.M.I. the statement $\dfrac{1}{3\cdot 6}+\dfrac{1}{6\cdot 9}+\dfrac{1}{9\cdot 12}+.........+\dfrac{1}{3n\cdot 3(n+1)}=\dfrac{n}{9\cdot (n+1)}$ is true.

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