Sum of Sequence of Fourth Powers

Theorem

 * $\displaystyle \sum_{k \mathop = 0}^n k^4 = \dfrac {\left({n + 1}\right) n \left({n + \frac 1 2}\right) \left({3 n^2 + 3 n - 1}\right)} {15}$

Also presented as
This result can also be presented as:


 * $\displaystyle \sum_{k \mathop = 0}^n k^4 = \dfrac {n \left({n + 1}\right) \left({2 n + 1}\right) \left({3 n^2 + 3 n - 1}\right)} {30}$