Six, (20 points) Find the summation formula for $\sum_{k=0}^{n} k^{4}$, and provide a proof.



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Six, from $(n+3)(n+2)(n+1) n$ $=n^{4}+6 n^{3}+11 n^{2}+6 n$,
we have $n^{4}=(n+3)(n+2)(n+1) n-6 n^{3}-11 n^{2}-6 n$.
Notice that
$$
\begin{array}{l}
\sum_{k=1}^{n}(k+3)(k+2)(k+1) k \\
=\sum_{k=1}^{n} \frac{1}{5}\{(k+3)(k+2)(k+1) k[(k+4)- \\
(k-1)]\} \\
=\frac{1}{5} \sum_{k=1}^{n}[(k+4)(k+3)(k+2)(k+1) k- \\
(k+3)(k+2)(k+1) k(k-1)] \\
=\frac{1}{5}(n+4)(n+3)(n+2)(n+1) n \text {. } \\
\end{array}
$$

We also know that $\sum_{k=1}^{n} k^{3}=\left(\sum_{k=1}^{n} k\right)^{2}=\frac{1}{4} n^{2}(n+1)^{2}$,
$$
\sum_{k=1}^{n} k^{2}=\frac{1}{6} n(n+1)(2 n+1), \sum_{k=1}^{n} k=\frac{1}{2} n(n+1) \text {. }
$$

From equation (1), we have
$$
\begin{array}{l}
\sum_{k=0}^{n} k^{4}=\sum_{k=1}^{n} k^{4} \\
=\sum_{k=1}^{n}(k+3)(k+2)(k+1) k- \\
6 \sum_{k=1}^{n} k^{3}-11 \sum_{k=1}^{n} k^{2}-6 \sum_{k=1}^{n} k \\
=\frac{1}{5}(n+4)(n+3)(n+2)(n+1) n-\frac{3}{2} n^{2} \text {. } \\
(n+1)^{2}-\frac{11}{6} n(n+1)(2 n+1)-3 n(n+1) \\
=\frac{1}{5}\left(n^{5}+10 n^{4}+35 n^{3}+50 n^{2}+24 n\right)- \\
\frac{3}{2}\left(n^{4}+2 n^{3}+n^{2}\right)-\frac{11}{6}\left(2 n^{3}+3 n^{2}+n\right)- \\
3 n^{2}-3 n \\
=\frac{1}{5} n^{5}+\frac{1}{2} n^{4}+\frac{1}{3} n^{3}-\frac{1}{30} n \text {. } \\
\end{array}
$$

Note: The main technique used in this solution is the "splitting terms for summation" method.
(Wang Xiaohui provided)