NCERT Solutions for Class 11 Maths Chapter 4 – Principle of Mathematical Induction
Chapter 4 is unlike any other chapter in the Class 11 syllabus. It does not introduce a new type of number or a new formula. Instead, it teaches you a proof technique — a rigorous, logical method for proving that a mathematical statement is true for all natural numbers. This technique is called the Principle of Mathematical Induction, and once you understand it, you will see it used everywhere from number theory to algorithm analysis in computer science.
The idea is remarkably intuitive once you grasp the domino analogy. Imagine an infinite row of dominoes. If you can prove that (1) the first domino falls, and (2) whenever any domino falls, the next one falls too — then you know all dominoes will eventually fall. Induction works on exactly this logic, applied to mathematical statements. For all Chapters must, read NCERT Solutions for Class 11 Maths and subject-wise NCERT Solutions for Class 11.
What makes this NCERT solutions chapter manageable is that every single question in Exercise 4.1 follows the same three-step structure: verify the base case, assume the statement for n = k, and prove it for n = k + 1. Once this template is drilled into muscle memory, even complex-looking induction problems become systematic. These solutions are written to show that structure explicitly in every problem, because CBSE awards step marks — you can score even if you make an error late in the proof, provided your method is clear.
Download PDF – All Exercises of NCERT Solutions for Class 11 Maths Chapter 4 Principle of Mathematical Induction
| Exercise | Topic Covered | Number of Questions |
|---|---|---|
| Exercise 4.1 | Proofs by Mathematical Induction (Sums, Divisibility, Inequalities) | 24 Questions |
| Miscellaneous | Advanced Induction Problems | 5 Questions |
Chapter 4 – Mathematical Induction: Concepts, Explanation and Key Tables
The Three Steps of Every Induction Proof
Mathematical Induction has an unbreakable structure. Every proof in NCERT follows these three steps, and CBSE expects all three to be written explicitly:
| Step | What to Do | What to Write |
|---|---|---|
| Step 1: Base Case | Verify the statement is true for n = 1 (or the smallest stated value) | Substitute n = 1 in both sides and show they are equal / statement holds |
| Step 2: Inductive Hypothesis | Assume the statement is true for n = k | Write "Let P(k) be true, i.e., [statement with n replaced by k]" |
| Step 3: Inductive Step | Using P(k), prove the statement for n = k + 1 | Add the (k+1)th term to both sides OR substitute (k+1) and simplify to reach LHS = RHS |
Missing any one of these steps in a CBSE answer results in partial marks at best.
Three Categories of Induction Problems in NCERT
| Category | How to Recognise It | Key Technique |
|---|---|---|
| Summation Formulas | Problem involves sum of a series (1² + 2² + ... + n²) | Add the (k+1)th term to P(k) and simplify RHS to match the formula with (k+1) substituted |
| Divisibility Statements | "Prove that n³ + 2n is divisible by 3" | Express P(k+1) using P(k); extract the assumed divisible expression |
| Inequality Statements | "Prove 2ⁿ > n for all n ≥ 1" | Multiply or add to P(k) and use properties of inequalities carefully |
Common Summation Results Proven by Induction
These appear directly in NCERT exercises — knowing the standard result before starting the induction saves you from making algebraic errors:
| Statement | Standard Result |
|---|---|
| 1 + 2 + 3 + ... + n | n(n+1)/2 |
| 1² + 2² + 3² + ... + n² | n(n+1)(2n+1)/6 |
| 1³ + 2³ + 3³ + ... + n³ | [n(n+1)/2]² |
| 1 + 3 + 5 + ... + (2n–1) | n² |
| a + (a+d) + ... + (a+(n–1)d) | n/2 × [2a + (n–1)d] |
| a + ar + ar² + ... + arⁿ⁻¹ | a(rⁿ – 1)/(r – 1), r ≠ 1 |
Divisibility Problems — The Trick That Always Works
For problems of the type "Prove 4ⁿ – 1 is divisible by 3", the standard approach is:
- Assume 4ᵏ – 1 = 3m for some integer m (this is P(k)).
- Write 4ᵏ⁺¹ – 1 = 4 × 4ᵏ – 1 = 4(3m + 1) – 1 = 12m + 3 = 3(4m + 1).
- Since (4m + 1) is an integer, the expression is divisible by 3.
The key move is always expressing 4ᵏ⁺¹ in terms of 4ᵏ, then substituting the inductive hypothesis.
Inequality Problems — Pitfalls to Avoid
| Common Error | What to Do Instead |
|---|---|
| Using ≥ and > interchangeably | Be precise — if P(k) says 2ᵏ ≥ k + 1, make sure your proof for P(k+1) uses that exact assumption |
| Trying to prove from scratch without using P(k) | Always use the inductive hypothesis — that is the entire point of induction |
| Skipping the base case because it seems obvious | Always write it — CBSE awards a separate mark for it |
| Writing "the result follows by induction" without completing Step 3 | Step 3 must be fully worked out — stating the conclusion without proof loses marks |
Study Tips for Chapter 4
- The inductive step is where most marks are won or lost. Practise expressing P(k+1) as P(k) plus one extra term — this is the core algebraic move.
- For divisibility, write the assumption in the form "expression = multiple × integer" explicitly. This makes Step 3 clean.
- This chapter is a short one (just Exercise 4.1) but the questions are long — practise writing complete proofs within 8–10 lines.
- The miscellaneous exercise has problems that mix induction with inequalities and complex expressions — attempt these after Exercise 4.1 is fully comfortable.