Higher June 2019 Paper 1 Q13
13 Given that \(n\) can be any integer such that \(n \gt 1\), prove that \(n^2 - n\) is never an odd number. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| Proof shown | C2 | for complete argument, eg \(n(n - 1)\) is the product of two consecutive integers and must be even as either \(n\) or \(n - 1\) must be even or gives correct reasoning for \(n\) odd and \(n\) even \(n\) odd: odd × odd = odd and odd − odd = even \(n\) even: even × even = even and even − even = even or \(n\) odd: \((2n + 1)^2 - (2n + 1) = 4n^2 + 2n = 2(2n^2 + n)\) \(n\) even: \((2n)^2 - (2n) = 4n^2 - 2n = 2(2n^2 - n)\) |
| (C1 | for factorising, eg \(n(n - 1)\) OR gives correct reasoning for \(n\) odd or \(n\) even OR gives a partial explanation using \(n\) odd and \(n\) even, eg odd2 − odd = even and even2 − even = even) |