Following from the question here, I was wondering if it's possible show directly that $$\sum_{r=1}^n r^4=\frac{3n^2+3n-1}5\sum_{r=1}^n r^2$$ without expanding the summation in full on either side.
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1Haven't you try to use induction by $n$? – Anton Grudkin Jun 09 '16 at 13:28
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I was hoping for a more direct approach. – Hypergeometricx Jun 09 '16 at 17:04
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We have that \begin{align}\sum_{r=1}^n r^2\sum_{r=1}^n r-\sum_{r=1}^n r^3 &=\sum_{1\le r< s\le n}rs^2+\sum_{1\le r<s\le n}sr^2\\ &=\sum_{r=1}^n r \frac{r(r-1)(2r-1)}{6}+\sum_{r=1}^n\frac{r(r-1)}{2}r^2\\ &=\frac56\sum_{r=1}^nr^4-\sum_{r=1}^nr^3+\frac1{6}\sum_{r=1}^nr^2,\end{align} where $\sum r^3$ cancels so that $$5\sum_{r=1}^nr^4=(6\sum_{r=1}^nr -1)\sum_{r=1}^n r^2=(3n^2+3n-1)\sum_{r=1}^n r^2.$$
ArtW
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1Brilliant solution! Exactly what I was looking for :) Thanks! (+1, and accepted) – Hypergeometricx Jun 09 '16 at 17:10
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If you know that $\sum_{r=1}^n r^4$ is a multiple of $\sum_{r=1}^n r^2$ as polynomials, then you can simply evaluate their quotient at three different points and reconstruct the polynomial $\frac{3n^2+3n-1}5$ by interpolation.
This assumes that you also know that $\sum_{r=1}^n r^k$ is a polynomial of degree $k+1$ in $n$.
lhf
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