mathschallenge.net

Problem

The sum of the first $n$ squares, $1^2 + 2^2 + ... + n^2 = \dfrac{n(n+1)(2n+1)}{6}$.

For example, $1^2 + 2^2 + ... + 10^2 = \dfrac{10 \times 11 \times 21}{6} = 385$.

Prove that $n(n+1)(2n+1)$ is divisible by six for all integer values, $n$.