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.