Sum of Sequence of Squares/Proof by Summation of Summations

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

 * [[File:Sum of Sequences of Squares.jpg]]

We can observe from the above diagram that:
 * $\displaystyle \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \sum_{i \mathop = 1}^n \left({\sum_{j \mathop = i}^n j}\right)$

Therefore we have: