Sum of Sequence of Squares/Proof by Products of Consecutive Integers

Theorem

 * $\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \left({n + 1}\right) \left({2 n + 1}\right)} 6$

Proof
Observe that:

This can then be used as the basis of a Telescoping Series, as follows:

Then: