Sum of Sequence of Squares/Also presented as

Theorem
The Sum of Sequence of Squares can also be presented as:
 * $\ds \forall n \in \N: \sum_{i \mathop = 0}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$

This is seen to be equivalent to the given form by the fact that the first term evaluates to $\dfrac {0 \paren {0 + 1} \paren {2 \times 0 + 1} } 6$ which is zero.