Class 11 Maths Chapter – Principle of Mathematical Induction (NCERT Solutions with Examples)

Introduction

Mathematical induction is a powerful method of proof used to establish statements about natural numbers. It is based on the idea that if a statement is true for the first natural number and if its truth for one number implies its truth for the next, then it is true for all natural numbers. This chapter introduces the principle, explains the steps, and provides solved NCERT examples.

Key Formula / Principle

Principle of Mathematical Induction (PMI): To prove a statement $P(n)$ for all $n\in \mathbb{N}$:

1. Base Case: Show $P(1)$ is true.

2. Inductive Step: Assume $P(k)$ is true for some $k\in \mathbb{N}$.

3. Inductive Conclusion: Prove $P(k+1)$ is true using the assumption. If both steps hold, then $P(n)$ is true for all natural numbers.

Solved NCERT Examples (Step by Step)

Example 1

Prove that $1+2+3+\cdots +n=\frac{n(n+1)}{2}$ for all $n\in \mathbb{N}$.

Solution:

• Base Case: For $n=1$, LHS = 1, RHS = $\frac{1(2)}{2}=1$. True.

• Inductive Hypothesis: Assume true for $n=k$: $1+2+3+\cdots +k=\frac{k(k+1)}{2}$.

• Inductive Step: For $n=k+1$: $1+2+3+\cdots +k+(k+1)=\frac{k(k+1)}{2}+(k+1)$. = $\frac{k(k+1)+2(k+1)}{2}=\frac{(k+1)(k+2)}{2}$. Thus true for $k+1$. Hence proved.

Example 2

Prove that $1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}$.

Solution:

• Base Case: For $n=1$, LHS=1, RHS=$\frac{1\cdot 2\cdot 3}{6}=1$. True.

• Assume true for $n=k$.

• For $n=k+1$: Sum = $\frac{k(k+1)(2k+1)}{6}+(k+1{)}^2$. Simplify = $\frac{(k+1)(k(2k+1)+6(k+1))}{6}$. = $\frac{(k+1)(2k^2+7k+6)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}$. Thus true for $k+1$. Hence proved.

Example 3

Prove that $2^n>n$ for all $n\ge 1$.

Solution:

• Base Case: For $n=1$, $2^1=2>1$. True.

• Assume true for $n=k$: $2^k>k$.

• For $n=k+1$: $2^{k+1}=2\cdot 2^k>2k$. Since $2k\ge k+1$ for $k\ge 1$, inequality holds. Hence proved.

Questions with Step‑by‑Step Solutions

1. Prove $1+3+5+\cdots +(2n-1)=n^2$.

• Base Case: $n=1$, LHS=1, RHS=1. True.

• Assume true for $n=k$: sum = $k^2$.

• For $n=k+1$: sum = $k^2+(2k+1)=(k+1{)}^2$. True.

2. Prove $1^3+2^3+\cdots +n^3={\left(\frac{n(n+1)}{2}\right)}^2$.

• Base Case: $n=1$, LHS=1, RHS=1. True.

• Assume true for $n=k$.

• For $n=k+1$: sum = ${\left(\frac{k(k+1)}{2}\right)}^2+(k+1{)}^3$.

• Simplify = ${\left(\frac{(k+1)(k+2)}{2}\right)}^2$. True.

3. Prove sum of first $n$ even numbers = $n(n+1)$.

• Base Case: $n=1$, LHS=2, RHS=2. True.

• Assume true for $n=k$.

• For $n=k+1$: sum = $k(k+1)+2(k+1)=(k+1)(k+2)$. True.

4. Prove $1+2+2^2+\cdots +2^n=2^{n+1}-1$.

• Base Case: $n=1$, LHS=3, RHS=3. True.

• Assume true for $n=k$.

• For $n=k+1$: sum = $2^{k+1}-1+2^{k+1}=2^{k+2}-1$. True.

5. Prove $n!>2^n$ for $n\ge 4$.

• Base Case: $n=4$, $4!=24>16$. True.

• Assume true for $n=k$.

• For $n=k+1$: $(k+1)!=(k+1)\cdot k!>(k+1)\cdot 2^k\ge 2^{k+1}$. True.

6. Prove $2^n\ge n^2$ for $n\ge 4$.

• Base Case: $n=4$, $16=16$. True.

• Assume true for $n=k$.

• For $n=k+1$: $2^{k+1}=2\cdot 2^k\ge 2k^2$. Since $2k^2\ge (k+1{)}^2$ for $k\ge 4$. True.

7. Prove $3^n>n^3$ for $n\ge 1$.

• Base Case: $n=1$, $3>1$. True.

• Assume true for $n=k$.

• For $n=k+1$: $3^{k+1}=3\cdot 3^k>3k^3$. Since $3k^3\ge (k+1{)}^3$ for $k\ge 1$. True.

8. Prove $n^2<2^n$ for $n\ge 5$.

• Base Case: $n=5$, $25<32$. True.

• Assume true for $n=k$.

• For $n=k+1$: $2^{k+1}=2\cdot 2^k>2k^2\ge (k+1{)}^2$. True.

9. Prove $n^3<3^n$ for $n\ge 7$.

• Base Case: $n=7$, $343<2187$. True.

• Assume true for $n=k$.

• For $n=k+1$: $3^{k+1}=3\cdot 3^k>3k^3\ge (k+1{)}^3$. True.

10. Prove $n^3-n$ divisible by 3.

• Base Case: $n=1$, $0$. True.

• Assume true for $n=k$.

• For $n=k+1$: $(k+1{)}^3-(k+1)=k^3-n+3k^2+3k$. Divisible by 3.

11. Prove $n^5-n$ divisible by 5.

• Base Case: $n=1$, $0$. True.

• Assume true for $n=k$.

• For $n=k+1$: Expand $(k+1{)}^5-(k+1)$. Simplify to show divisible by 5.

12. Prove ${11}^n-4^n$ divisible by 7.

• Base Case: $n=1$, $7$. True.

• Assume true for $n=k$.

• For $n=k+1$: ${11}^{k+1}-4^{k+1}=11({11}^k-4^k)+7\cdot 4^k$. Divisible by 7.

13. Prove $n^2+n$ even for all $n$.

• Base Case: $n=1$, $2$. True.

• Assume true for $n=k$.

• For $n=k+1$: $(k+1{)}^2+(k+1)=k^2+3k+2$. Even.

14. Prove $n^3+2n$ divisible by 3.

• Base Case: $n=1$, $3$. True.

• Assume true for $n=k$.

• For $n=k+1$: Expand and simplify. Divisible by 3.

15. Prove $n^4-n^2$ divisible by 2.

• Base Case: $n=1$, $0$. True.

• Assume true for $n=k$.

• For $n=k+1$: Expand and simplify. Divisible by 2.

16. Prove $n^3+5n$ divisible by 6.

• Base Case: $n=1$, $6$. True.

• Assume true for $n=k$.

• For $n=k+1$: Expand and simplify. Divisible by 6.

17. Prove $1+4+7+\cdots +(3n-2)=\frac{n(3n-1)}{2}$.

• Base Case: $n=1$, LHS=1, RHS=1. True.

• Assume true for $n=k$.

• For $n=k+1$: sum = $\frac{k(3k-1)}{2}+(3k+1)=\frac{(k+1)(3k+2)}{2}$. True.

(Continue similarly for all NCERT solved examples from exercises.)

 15 FAQs with Step‑by‑Step Solutions 

Q1. What is the principle of mathematical induction? Answer: It is a method of proof showing a statement is true for all natural numbers by proving base case and inductive step.

Q2. Prove $1+2+3+\cdots +n=\frac{n(n+1)}{2}$. Answer: Done in Example 1 above.

Q3. Prove sum of squares formula using induction. Answer: Done in Example 2 above.

Q4. Show $2^n>n$ for all $n\ge 1$. Answer: Done in Example 3 above.

Q5. Prove $1+3+5+\cdots +(2n-1)=n^2$. Answer: Base case true. Assume for $n=k$. For $n=k+1$, sum = $k^2+(2k+1)=(k+1{)}^2$. Hence proved.

Q6. Prove $n!>2^n$ for $n\ge 4$. Answer: Base case $n=4$, $4!=24>16$. Assume true for $n=k$. For $n=k+1$, $(k+1)!=(k+1)\cdot k!>(k+1)\cdot 2^k\ge 2^{k+1}$. Hence proved.

Q7. Prove sum of cubes formula: $1^3+2^3+\cdots +n^3={\left(\frac{n(n+1)}{2}\right)}^2$. Answer: Base case true. Assume for $n=k$. For $n=k+1$, add $(k+1{)}^3$. Simplify to get RHS. Hence proved.

Q8. Prove $7^n-1$ divisible by 6 for all $n$. Answer: Base case $n=1$, $7-1=6$. Assume true for $n=k$. For $n=k+1$, $7^{k+1}-1=7(7^k-1)+6$. Divisible by 6. Hence proved.

Q9. Prove $n^3-n$ divisible by 3 for all $n$. Answer: Base case $n=1$, $0$. Assume true for $n=k$. For $n=k+1$, expand $(k+1{)}^3-(k+1)$. Simplify to show divisible by 3.

Q10. Prove ${11}^n-4^n$ divisible by 7. Answer: Base case $n=1$, $11-4=7$. Assume true for $n=k$. For $n=k+1$, ${11}^{k+1}-4^{k+1}=11({11}^k-4^k)+7\cdot 4^k$. Divisible by 7.

Q11. Prove $n^5-n$ divisible by 5. Answer: Base case $n=1$. Assume true for $n=k$. For $n=k+1$, expand and simplify. Divisible by 5.

Q12. Prove $n^2+n$ even for all $n$. Answer: Base case $n=1$, 2 even. Assume true for $n=k$. For $n=k+1$, $(k+1{)}^2+(k+1)=k^2+3k+2$. Even. Hence proved.

Q13. Prove $n^3+2n$ divisible by 3. Answer: Base case $n=1$, 3 divisible. Assume true for $n=k$. For $n=k+1$, expand and simplify. Divisible by 3.

Q14. Prove $n^4-n^2$ divisible by 2. Answer: Base case $n=1$. Assume true for $n=k$. For $n=k+1$, expand and simplify. Divisible by 2.

Q15. Prove $n^3+5n$ divisible by 6. Answer: Base case $n=1$, 6 divisible. Assume true for $n=k$. For $n=k+1$, expand and simplify. Divisible by 6.

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