QCE Vault / Specialist Maths Proof by mathematical induction — Question 2 QCAA 2023, Paper 1 · 1 mark
Browse all questions Original practice exam Report an issue Q2 · 2023 · Technology-free Simple familiar 1 mark
QUESTION 2 Consider the proof of the following proposition using mathematical induction. ∑ r = 1 n r ( r + 1 ) = 1 3 n ( n + 1 ) ( n + 2 ) ∀ n ∈ Z + . \sum_{r=1}^{n} r(r+1)=\frac13n(n+1)(n+2)\quad\forall n\in\mathbb Z^+. r = 1 ∑ n r ( r + 1 ) = 3 1 n ( n + 1 ) ( n + 2 ) ∀ n ∈ Z + . An appropriate assumption statement within the proof is (A) ∑ r = 1 k k ( k + 1 ) = 1 3 k ( k + 1 ) ( k + 2 ) \displaystyle\sum_{r=1}^{k} k(k+1)=\frac13k(k+1)(k+2) r = 1 ∑ k k ( k + 1 ) = 3 1 k ( k + 1 ) ( k + 2 ) (B) ∑ r = 1 k k ( k + 1 ) = 1 3 n ( n + 1 ) ( n + 2 ) \displaystyle\sum_{r=1}^{k} k(k+1)=\frac13n(n+1)(n+2) r = 1 ∑ k k ( k + 1 ) = 3 1 n ( n + 1 ) ( n + 2 ) (C) ∑ r = 1 k r ( r + 1 ) = 1 3 k ( k + 1 ) ( k + 2 ) \displaystyle\sum_{r=1}^{k} r(r+1)=\frac13k(k+1)(k+2) r = 1 ∑ k r ( r + 1 ) = 3 1 k ( k + 1 ) ( k + 2 ) (D) ∑ r = 1 k r ( r + 1 ) = 1 3 n ( n + 1 ) ( n + 2 ) \displaystyle\sum_{r=1}^{k} r(r+1)=\frac13n(n+1)(n+2) r = 1 ∑ k r ( r + 1 ) = 3 1 n ( n + 1 ) ( n + 2 ) WORKED SOLUTION
Answer C 1 mark Worked solution
For the inductive hypothesis, replace n n n by k k k everywhere in the proposition while retaining the summation variable r r r : ∑ r = 1 k r ( r + 1 ) = 1 3 k ( k + 1 ) ( k + 2 ) \sum_{r=1}^{k}r(r+1)=\frac13k(k+1)(k+2) ∑ r = 1 k r ( r + 1 ) = 3 1 k ( k + 1 ) ( k + 2 ) . Worked explanation by QCE Vault; correct option verified against the QCAA answer key.
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