Sum of Sequence of Squares/Proof by Telescoping Series

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:
 * $3 i \left({i + 1}\right) = i \left({i + 1}\right) \left({i + 2}\right) - i \left({i + 1}\right) \left({i - 1}\right)$

That is:
 * $(1): \quad 6 T_i = \left({i + 1}\right) \left({\left({i + 1}\right) + 1}\right) \left({\left({i + 1}\right) - 1}\right) - i \left({i + 1}\right) \left({i - 1}\right)$

Then:

where $T_n$ is the $n$th Triangular number.

Then: