Use mathematical induction to prove that \(n({n^2} + 5)\) is divisible by 6 for \(n \in {\mathbb{Z}^ + }\).

let \({\text{P}}(n)\) be the proposition that \(n({n^2} + 5)\) is divisible by 6 for \(n \in {\mathbb{Z}^ + }\)

consider P(1):

when \(n = 1,{\text{ }}n({n^2} + 5) = 1 \times ({1^2} + 5) = 6\) and so P(1) is true     R1

assume \({\text{P}}(k)\) is true ie, \(k({k^2} + 5) = 6m\) where \(k,{\text{ }}m \in {\mathbb{Z}^ + }\)     M1

Note:     Do not award M1 for statements such as “let \(n = k\)”.

consider \({\text{P}}(k + 1)\):

\((k + 1)\left( {{{(k + 1)}^2} + 5} \right)\)    M1

\( = (k + 1)({k^2} + 2k + 6)\)

\( = {k^3} + 3{k^2} + 8k + 6\)    (A1)

\( = ({k^3} + 5k) + (3{k^2} + 3k + 6)\)    A1

\( = k({k^2} + 5) + 3k(k + 1) + 6\)    A1

\(k(k + 1)\) is even hence all three terms are divisible by 6     R1

\({\text{P}}(k + 1)\) is true whenever \({\text{P}}(k)\) is true and P(1) is true, so \({\text{P}}(n)\) is true for \(n \in {\mathbb{Z}^ + }\)     R1

Note:     To obtain the final R1, four of the previous marks must have been awarded.

[8 marks]

This proved to be a good discriminator. The average candidate seemed able to work towards \({\text{P}}(k + 1) = {k^3} + 3{k^2} + 8k + 6\), and a number made some further progress.

Unfortunately, even otherwise good candidates are still writing down incorrect or incomplete induction statements, such as ‘Let \(n = k\)’ rather than ‘Suppose true for \(n = k\)’ (or equivalent).

It was also noted than an increasing number of candidates this session assumed ‘\({\text{P}}(n)\) to be true’ before going to consider \({\text{P}}(n + 1)\). Showing a lack of understanding of the induction argument, these approaches scored very few marks.