Sum of Sequence of Squares

Theorem

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

Proof by Telescoping Sum
Observe that $3 i \left({i + 1}\right) = i \left({i + 1}\right) \left({i + 2}\right) - i \left({i + 1}\right) \left({i - 1}\right)$.

By taking the sum we'll get a telescoping one on the RHS and the conclusion follows.

Direct Proof

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

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

Therefore we have:

Historical Note
This result was documented by Āryabhaṭa in his work Āryabhaṭīya of 499 CE.