A formula for $1^4+2^4+...+n^4$

Question

I know that $$\sum^n_{i=1}i^2=\frac{1}{6}n(n+1)(2n+1)$$ and $$\sum^n_{i=1}i^3=\left(\sum^n_{i=1}i\right)^2.$$ Here is the question: is there a formula for $$\sum^n_{i=1}i^4.$$

Answer 1

We can get the formula by the following way.

$$(n+1)^5-1=\sum_{k=1}^n((k+1)^5-k^5)=\sum_{k=1}^n(5k^4+10k^3+10k^2+5k+1).$$ Thus, $$\sum_{k=1}^nk^4=$$ $$=\frac{1}{5}\left((n+1)^5-1-10\cdot\frac{n^2(n+1)^2}{4}-10\cdot\frac{n(n+1)(2n+1)}{6}-5\cdot\frac{n(n+1)}{2}-n\right)=$$ $$=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}.$$

Answer 2

yes there is a formula, $$\sum_{i=1}^ni^4=1/30\,n \left( 2\,n+1 \right) \left( n+1 \right) \left( 3\,{n}^{2}+3 \,n-1 \right) $$

Answer 3

It is interesting to note that $$\begin{align} \sum_{i=1}^n i^{\,4} &=\frac 1{30}n(n+1)(2n+1)(3n^2+3n-1)\\ &=\frac{3n^2+3n-1}5\cdot\frac {n(n+1)(2n+1)}6 \\ &=\frac{3n^2+3n-1}5\sum_{i=1}^n i^{\,2}\end{align}$$