Mathematical Induction: JEE Main Complete Guide

Mathematical induction is a proof technique that appears in JEE Main as one to two questions per session, typically in a straightforward format: verify the inductive step for a given statement. The chapter rewards students who understand the logical structure — base case, inductive hypothesis, inductive step — rather than those who mechanically copy a format without understanding it.

The Principle of Mathematical Induction

Principle of mathematical induction (PMI): Let P(n) be a statement involving the natural number n. If (1) P(1) is true [base case], and (2) whenever P(k) is true for some k ≥ 1, P(k+1) is also true [inductive step], then P(n) is true for all natural numbers n. The logic: the base case provides the starting point; the inductive step provides the "domino rule" — each true case implies the next. Together, they guarantee truth for all n.

JEE Main tests induction in three forms: (1) summation formulas — prove that 1 + 2 + 3 + ... + n = n(n+1)/2; (2) divisibility — prove that n³ − n is divisible by 6 for all n; (3) inequality — prove that 2ⁿ > n² for n ≥ 5. For each type, the structure of the proof is identical. The skill JEE Main tests is correctly computing P(k+1) from P(k) in the inductive step. For the sequences and series results that induction proves, see our Sequences and Series Guide.

Standard Summation Formula Proofs

Proof template for Σn = n(n+1)/2: P(n): 1+2+...+n = n(n+1)/2. Step 1 — P(1): LHS = 1; RHS = 1×2/2 = 1. ✓ Step 2 — Assume P(k): 1+2+...+k = k(k+1)/2. Step 3 — Prove P(k+1): 1+2+...+k+(k+1) = k(k+1)/2 + (k+1) = (k+1)[k/2 + 1] = (k+1)(k+2)/2. ✓ This is P(k+1). Hence by PMI, P(n) holds for all n ∈ ℕ. JEE Main replaces this with Σn² = n(n+1)(2n+1)/6, Σn³ = [n(n+1)/2]², or similar — the proof structure is identical, only the algebra differs. Take a free mock test on algebra and mathematical induction to practise these proof problems.

Divisibility Proofs

Prove 6 | (n³−n) for all n ∈ ℕ. Note: n³−n = n(n−1)(n+1) = product of three consecutive integers, always divisible by 3! = 6. But the PMI proof: P(1): 1³−1 = 0, divisible by 6 ✓. Assume P(k): 6 | (k³−k). P(k+1): (k+1)³−(k+1) = k³+3k²+3k+1−k−1 = (k³−k)+3k(k+1). 6 | (k³−k) by hypothesis; 6 | 3k(k+1) because k(k+1) is always even (product of consecutive integers), so 3k(k+1) is divisible by 6. Therefore 6 | [(k³−k)+3k(k+1)] ✓.

Inequality Proofs

Prove 2ⁿ > n² for n ≥ 5. Base case P(5): 2⁵ = 32 > 25 = 5². ✓ Assume P(k): 2ᵏ > k² for some k ≥ 5. P(k+1): 2^(k+1) = 2·2ᵏ > 2k² (by hypothesis). Need to show 2k² > (k+1)² = k²+2k+1. This requires k² > 2k+1, or k²−2k−1 > 0, i.e. (k−1)² > 2. This holds for k ≥ 3, so certainly for k ≥ 5. Therefore 2^(k+1) > 2k² > (k+1)² ✓. JEE Main tests inequality induction in the integer-answer section — the question usually gives a specific inequality and asks for the smallest n for which it holds, or asks you to identify the missing step in a given proof.

Strong Induction

Strong induction: assume P(1), P(2), ..., P(k) are all true and prove P(k+1). Used when the inductive step requires not just P(k) but several previous cases. Example: Fibonacci sequence properties. JEE Main rarely tests strong induction directly but may ask about it conceptually.

Exam Strategy

JEE Main induction questions are template-driven. In MCQ format: you are usually asked to verify the inductive step expression — expand P(k+1) directly from the statement and simplify, then match to the options. In integer format: compute a specific sum or prove a specific divisibility result. The critical skill is algebraic expansion accuracy in the inductive step. Practise expanding P(k+1) from P(k) for ten different statements until you can do it without errors. For the algebra of sequences underlying induction, see our Geometric Progressions Guide. Upgrade for ₹149/month for 60+ mathematical induction problems at all JEE difficulty levels.