Prove by mathematical induction that \({n^3} + 11n\) is divisible by 3 for all \(n \in {\mathbb{Z}^ + }\).

\(n = 1:{\text{ }}{1^3} + 11 = 12\)

\( = 3 \times 4\) or a multiple of 3     A1

assume the proposition is true for \(n = k\)   \((ie{\text{ }}{k^3} + 11k = 3{\text{ m)}}\)     M1

Note:     Do not award M1 for statements with “Let \(n = k\)”.

consider \(n = k + 1:\)   \({(k + 1)^3} + 11(k + 1)\)     M1

\( = {k^3} + 3{k^2} + 3k + 1 + 11k + 11\)     A1

\( = {k^3} + 11k + (3{k^2} + 3k + 12)\)     M1

\( = 3(m + {k^2} + k + 4)\)     A1

Note:     Accept \({k^3} + 11k + 3({k^2} + k + 4)\) or statement that \({k^3} + 11k + (3{k^2} + 3k + 12)\) is a multiple of 3.

true for \(n = 1\), and \(n = k{\text{ true }} \Rightarrow n = k + 1{\text{ true}}\)

hence true for all \(n \in {\mathbb{Z}^ + }\)     R1

Note:     Only award the final R1 if at least 4 of the previous marks have been achieved.

[7 marks]

It was pleasing to see a great many clear and comprehensive answers for this relatively straightforward induction question. The inductive step only seemed to pose problems for the very weakest candidates. As in previous sessions, marks were mainly lost by candidates writing variations on ‘Let \(n = k\)’, rather than ‘Assume true for \(n = k\)’. The final reasoning step still needs attention, with variations on ‘\(n = k + 1{\text{ true }} \Rightarrow n = k{\text{ true}}\)’ evident, suggesting that mathematical induction as a technique is not clearly understood.