Sum of Sequence of Squares/Proof by Sum of Differences of Cubes

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
On the other hand:

Therefore:

Therefore: