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Question
Using mathematical induction, the expression n3−nn^3 - n is divisible by 6 for all n∈Nn \in \mathbb{N}. In the inductive step, (k+1)3−(k+1)−(k3−k)(k+1)^3 - (k+1) - (k^3 - k) equals:
(A)3k2+3k3k^2 + 3k
(B)3k2+3k+23k^2 + 3k + 2
(C)3k(k+1)+23k(k+1) + 2
(D)k2+kk^2 + k
Solution Path
Expand (k+1)3−(k+1)(k+1)^3 - (k+1), subtract k3−kk^3 - k to get 3k2+3k=3k(k+1)3k^2 + 3k = 3k(k+1). Product of consecutive integers is even, so this is divisible by 6.
01Question Setup
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Prove n3−nn^3 - n is divisible by 6 for all n∈Nn \in \mathbb{N} using induction. Find what (k+1)3−(k+1)−(k3−k)(k+1)^3 - (k+1) - (k^3 - k) simplifies to.
Find the difference between consecutive terms
02Algebraic Expansion
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Expand (k+1)3−(k+1)=k3+3k2+3k+1−k−1=k3+3k2+2k(k+1)^3 - (k+1) = k^3 + 3k^2 + 3k + 1 - k - 1 = k^3 + 3k^2 + 2k. Then subtract k3−kk^3 - k to get the difference.
(k+1)3−(k+1)=k3+3k2+2k(k+1)^3 - (k+1) = k^3 + 3k^2 + 2k
03Difference SimplificationKEY INSIGHT
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(k3+3k2+2k)−(k3−k)=3k2+3k=3k(k+1)(k^3 + 3k^2 + 2k) - (k^3 - k) = 3k^2 + 3k = 3k(k+1). Since k(k+1)k(k+1) is always even (product of consecutive integers), 3k(k+1)3k(k+1) is divisible by 6.
3k2+3k=3k(k+1)3k^2 + 3k = 3k(k+1), divisible by 6
04Final Answer
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The difference is 3k2+3k3k^2 + 3k, confirming that if k3−kk^3 - k is divisible by 6, so is (k+1)3−(k+1)(k+1)^3 - (k+1).
3k2+3k\boxed{3k^2 + 3k} - Answer (A)
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Principle of Mathematical Induction

STANDARD

To prove P(n) for all natural numbers: (1) Base case: show P(1) is true. (2) Inductive step: assume P(k) is true, then prove P(k+1) is true. Both steps together establish P(n) for all n.

P(1) true +[P(k)⇒P(k+1)]  ⟹  P(n) true ∀ n≥1P(1) \text{ true } + [P(k) \Rightarrow P(k+1)] \implies P(n) \text{ true } \forall \, n \geq 1

Induction questions in JEE typically ask students to identify the correct inductive step or find where an induction proof breaks down. Understanding the structure prevents errors.

Divisibility proofsSummation identitiesInequality proofs
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