A2 June 2019 Paper 2 Q10

Typical solution

Let \(n = 1\) then \(\mathrm{f}(1) = 12 = 2 \times 6\)

so the result is true for \(n = 1\)

Assume the result is true for \(n = k\):

Then \(\mathrm{f}(k) = 6m\) for some integer \(m\)

\[\begin{aligned}\mathrm{f}(k + 1) &= (k + 1)^3 + 3(k + 1)^2 + 8(k + 1) \\ &= k^3 + 3k^2 + 3k + 1 + 3(k^2 + 2k + 1) + 8(k + 1) \\ &= k^3 + 6k^2 + 17k + 12\end{aligned}\]\[\begin{aligned}\mathrm{f}(k + 1) &= k^3 + 6k^2 + 17k + 12 - (k^3 + 3k^2 + 8k) + \mathrm{f}(k) \\ &= 6m + 3k^2 + 9k + 12\end{aligned}\]

\(3k^2 + 9k + 12\) is a multiple of 3

\(3k^2 + 9k + 12 = 3(k^2 + 3k) + 12\)

\(k^2 + 3k = k(k + 3)\) and one of \(k\) or \(k + 3\) is even

\(\therefore k^2 + 3k\) is even and \(3(k^2 + 3k + 4)\) is an even multiple of 3 and hence divisible by 6

\(\therefore \mathrm{f}(k + 1)\) is divisible by 6 if \(\mathrm{f}(k)\) is divisible by 6

We also know that \(\mathrm{f}(1)\) is divisible by 6, so by induction this completes the proof.

Notes

(corrected from the printed mark scheme: the printed line reads \(\mathrm{f}(k + 1) = k^3 + 6k^2 + 17k + 12 - (k^3 + 3k^2 + 8k)\); the \(+\,\mathrm{f}(k)\) term, which gives the \(6m\), was missing.)