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So we have the following: $1^4 + 2^4 + 3^4 + ... + n^4$
How do you find a polynomial formula for this recursive relation? My attempt is to set it up as following: $(n+1)(n^3+1)$ but it does not look right.

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Note that we have $$(k+1)^5 - k^5 = 5k^4 + 10k^3 + 10k^2 + 5k + 1$$ Now sum this from $k=1$ to $n$ to get that $$\sum_{k=1}^n \left((k+1)^5 - k^5 \right) = 5\sum_{k=1}^n k^4 + 10\sum_{k=1}^n k^3 + 10\sum_{k=1}^n k^2 + 5\sum_{k=1}^n k + \sum_{k=1}^n 1$$ The terms on the left hand side can be simplified by telescopic cancellation to get \begin{align} (n+1)^5 - 1 & = 5\sum_{k=1}^n k^4 + 10\sum_{k=1}^n k^3 + 10\sum_{k=1}^n k^2 + 5\sum_{k=1}^n k + \sum_{k=1}^n 1 \tag{$\star$} \end{align} Now make use of the fact that \begin{align} \sum_{k=1}^n k^3 & = \dfrac{n^2(n+1)^2}4\\ \sum_{k=1}^n k^2 & = \dfrac{n(n+1)(2n+1)}6\\ \sum_{k=1}^n k & = \dfrac{n(n+1)}2\\ \sum_{k=1}^n 1 & = n \end{align} in $(\star)$ to get the value of $\displaystyle \sum_{k=1}^n k^4$. If you are not aware of the above relations you can micmic the procedure we used to compute $\displaystyle \sum_{k=1}^n k^4$, to compute these as well.

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I'm not the one who asked this, but I really like this proof. Clever! – Suugaku Apr 5 '13 at 3:16

Note that the polynomial $p$ you desire will have the properties that $p(0)=0$ and $p(n+1)-p(n)=(n+1)^4$ for all (nonnegative) integers $n$. That turns out to be enough to completely characterize the polynomial. To narrow it down a bit further for you, though, it will be a degree $5$ polynomial, with $\frac15$ as leading coefficient.

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