TITLE: Closed formula for linear binomial identity
QUESTION [6 upvotes]: I have the following identity:
\begin{equation}
m^4 = Z{m\choose 4}+Y{m\choose 3}+X{m\choose 2}+W{m\choose 1}
\end{equation}
I solved for the values and learned of the interpretation of W, X, Y, and Z in my last post:
Combinatorial reasoning for linear binomial identity
Now, I am interested in using the above to find a closed form solution for:
\begin{equation}
\sum\limits_{k=1}^nk^4 \\
\end{equation}
I do not see how to get a closed form solution to the above sum. What steps can I take? I especially would like to use the work from my last post.
Thanks!

REPLY [3 votes]: Use the identity
$$
\sum_{j=0}^{k-m}\binom{k-j}{m}\binom{j}{n}=\binom{k+1}{m+n+1}
$$
setting $m=0$ to get
$$
\sum_{j=0}^{k}\binom{j}{n}=\binom{k+1}{n+1}
$$
Then, once we have that
$$
m^4=24\binom{m}{4}+36\binom{m}{3}+14\binom{m}{2}+1\binom{m}{1}+0\binom{m}{0}
$$
we get that
$$
\begin{align}
\sum_{m=1}^k m^4
&=24\binom{k+1}{5}+36\binom{k+1}{4}+14\binom{k+1}{3}+1\binom{k+1}{2}+0\binom{k+1}{1}\\
&=24\binom{k}{5}+60\binom{k}{4}+50\binom{k}{3}+15\binom{k}{2}+1\binom{k}{1}+0\binom{k}{0}
\end{align}
$$