TMUA-P2-PRF1-011 — Paper P2 · Direct proof · Level 3

Options

1. A
   If $n = 2k$ then $n^2+n = 4k^2+2k = 2(2k^2+k)$, even; if $n = 2k+1$ then $n^2+n = (4k^2+4k+1)+(2k+1) = 4k^2+6k+2 = 2(2k^2+3k+1)$, even. Both cases give an even number.
2. B
   Check $n = 4$: $4^2+4 = 20$, which is even, so the claim holds for all integers $n$.
3. C
   If $n^2+n$ is even then $n(n+1)$ is even, so $n$ or $n+1$ is even, which proves the claim.
4. D
   If $n = 2k+1$ then $n^2+n = 4k^2+4k+1 = 2(2k^2+2k)+1$, which is odd, so the claim is false.

Letters only — the answer key is hidden. Work it out, then check the solution below.

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