<s>[INST] How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$? How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$?
I've spent quite some time on this problem. So far, I've simplified the right-hand side to $\frac{1}{30}(n+1)[(2n+3)(3n^3)+n(n-1)]$. But then, the algebra becomes very complicated when I add $(n+1)^4$ to both sides of the inductive hypothesis.
 [/INST] We have 
\begin{eqnarray*}
1^4+\cdots + n^4 &=& \frac{1}{5} n^5+\frac{1}{2}n^4+\frac{1}{3}n^3+\frac{1}{30}n) \\
&=& \frac{1}{30}(6 n^5+15n^4+10n^3-n). \\
\end{eqnarray*}
So
\begin{eqnarray*}
1^4+\cdots + n^4 +(n+1)^4 &=& \frac{1}{30}(6 n^5+15n^4+10n^3-n +30n^4+120n^3+180n^2+120n+30). \\
\end{eqnarray*}
Now check this really does equal the expression below
\begin{eqnarray*}
 \frac{1}{30}(6 (n^5 +5n^4+10n^3+10n^2+5n+1) \\
 +15(n^4+4n^3+6n^2+4n+1) \\
 +10(n^3+3n^2+3n+1) \\
 -(n+1) ). \\
\end{eqnarray*}
So 
\begin{eqnarray*}
1^4+\cdots + n^4 +(n+1)^4 &=& \frac{1}{30}(6 (n+1)^5+15(n+1)^4+10(n+1)^3-(n+1)). \\
\end{eqnarray*}
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