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

Last updated date: 02nd Aug 2024
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## CBSE Class 11 Maths Chapter-4 Important Questions - Free PDF Download

Free PDF download of Important Questions with solutions for CBSE Class 11 Maths Chapter 4 - Principle of Mathematical Induction prepared by expert Maths teachers from latest edition of CBSE(NCERT) books. Register online for Maths tuition on Vedantu.com to score more marks in your Examination.

Also, check CBSE Class 11 Maths Important Questions for other chapters:

 CBSE Class 11 Maths Important Questions Sl.No Chapter No Chapter Name 1 Chapter 1 Sets 2 Chapter 2 Relations and Functions 3 Chapter 3 Trigonometric Functions 4 Chapter 4 Principle of Mathematical Induction 5 Chapter 5 Complex Numbers and Quadratic Equations 6 Chapter 6 Linear Inequalities 7 Chapter 7 Permutations and Combinations 8 Chapter 8 Binomial Theorem 9 Chapter 9 Sequences and Series 10 Chapter 10 Straight Lines 11 Chapter 11 Conic Sections 12 Chapter 12 Introduction to Three Dimensional Geometry 13 Chapter 13 Limits and Derivatives 14 Chapter 14 Mathematical Reasoning 15 Chapter 15 Statistics 16 Chapter 16 Probability
Competitive Exams after 12th Science

## Study Important Questions for 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.

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

1. What is meant by Mathematical Induction in Chapter 4 of Class 11th Maths?

Mathematical induction Chapter 4  Class 11th Maths is a specialized way of proving specific mathematical statements accepted in algebra and other fields of applied maths, like inductive and deductive reasoning. If you visit the website of Vedantu, you will find the important questions that explain the concept of mathematical induction in a simple way. If you go through them and practice it well, you will be able to score full marks. These are required to practice to have a strong foundation too, which will help you in higher studies.

2. What are the two steps involved in the principles of mathematical induction?

The two steps that are involved in the principles of mathematical induction are as follows:

• In the first step, you need to prove that the statement given in the question is true for the initial value. This step is often called the base step.

• In the second step, you need to prove that the statement given in the question holds for the nth value as well as that for the (n+1)th iteration too.

3. How do you solve a math induction question in Chapter 4 of Class 11th Maths?

Through mathematical induction, you can prove that a statement, theorem or formula is true for every natural number n. For this, you must prove that the statement, theorem or formula is true for n=1 and that for n=kth number. It will show that it is true for any other value of n too. For example, if we want to know whether the formula of the sum of positive natural numbers, i.e. 1 + 2 + 3 … … … n = [n(n+1)/2] holds for all natural numbers, we first check with the smallest possible value of n, and then with higher numbers. If it is valid for both, we accept that it is valid for any natural number. The notes and solutions of Chapter 4 of Class 11th Maths are present on Vedantu's official website (vedantu.com) and mobile app for free of cost.

4. Who invented mathematical induction which was discussed in Chapter 4 of Class 11th Maths?

Pascal gave the formulas of mathematical induction for the first time in his book Traité du triangle arithmétique in 1665. But much before him, Gersonides is said to have used the principles of mathematical induction vigorously. Using the concepts of mathematical induction, we can check whether a statement, theorem or formula holds for any value of 'n', that is for any natural number, or not. We do it in two main steps, basic and inductive.

5. How can you prove the principle of mathematical induction in Chapter 4 of Class 11th Maths?

At first, we need to assume an initial value for 'n', and prove that the statement given in the question is valid for this initial value of n. Then, we need to assume another value of n, e.g. n=k, and prove that the given statement is true for this value of n too. After that, we need to do this process assuming n= k+1. In the end, we need to segregate n=k+1 into two parts and prove that the statement is valid for both parts.