Sum of Sequence of Fourth Powers

Theorem

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

Also presented as
This result can also be presented as:


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