## Expressing Divisibility

#### 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, $n$.

#### Solution

To prove that $n(n+1)(2n+1)$ is divisible by 6 we must show that it is divisible by both 2 and 3.

As $n$ and $n+1$ are consecutive integers then exactly one of them will be even and thus divisible by 2.

If neither $n$ nor $n+1$ is divisible by 3 then $n+2$ will be. Therefore $2(n+2) = 2n + 4$ and $2n + 4 -3 = 2n + 1$ will be divisible by 3.

Hence $n(n+1)(2n+1)$ is divisible by both 2 and 3. Q.E.D.

Problem ID: 356 (26 Aug 2009)     Difficulty: 2 Star

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