<s>[INST] A formula for $1^4+2^4+...+n^4$ 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 
$$\sum^n_{i=1}i^4.$$
 [/INST] 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,
$$\sum_{k=1}^nk^4=$$
$$=\frac{1}{5}\left((n+1)^5-1-10\cdot\frac{n^2(n+1)^2}{4}-10\cdot\frac{n(n+1)(2n+1)}{6}-5\cdot\frac{n(n+1)}{2}-n\right)=$$
$$=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}.$$
</s>