Proof by Mathematical Induction

00:00 6 marks

Use mathematical induction to prove that \(\sum_{r=1}^{n} r(r+1) = \frac{n(n+1)(n+2)}{3}\) for \(n \in \mathbb{Z}^{+}\).

Hint Hint

Hint

Verify the base case for n = 1, assume the formula holds for n = k, then show that it holds for n = k+1 by adding the term \((k+1)(k+2)\) and simplifying.

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Complete Answer

Base Case: For n = 1, the left-hand side is 1×(1+1) = 2 and the right-hand side is (1×2×3)/3 = 2. Induction Hypothesis: Assume for some k ∈ ℤ⁺ that \(\sum_{r=1}^{k} r(r+1) = \frac{k(k+1)(k+2)}{3}\). Induction Step: For n = k + 1, add \((k+1)(k+2)\) to both sides to obtain \(\frac{k(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}\). Thus, by mathematical induction, the formula holds for all positive integers.

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