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The sum of the fourth powers of the first $n$ integers can be expressed as a multiple of the sum of squares of the first $n$ integers, i.e.

$$\begin{align} \sum_{r=1}^n r^4&=\frac {n(n+1)(2n+1)(3n^3+3n-1)}{30}\\ &=\frac{3n^2+3n-1}5\cdot \frac {n(n+1)(2n+1)}6 \\ &=\frac{3n^2+3n-1}5\sum_{r=1}^nr^2 \end{align}$$

Question: Is it possible to show this, purely by manipulating the summand, and without first expressing the summation in closed form and then factoring the sum of squares?

From the question here, we see that

$$\sum_{r=1}^n r^4=\left(\sum_{r=1}^n r^2\right)^2-2\sum_{r=1}^n r^2\sum_{j=1}^{r-1}j^2\\ =\sum_{r=1}^n r^2 \left(\sum_{i=1}^n i^2-2\sum_{j=1}^{r-1}j^2\right)$$ but this does not appear to lead anywhere closer to answering the question.

Another approach might be to use Abel's summation formula $$\sum_{r=1}^n f_r (g_{r+1}-g_r)=\left[f_{n+1}g_{n+1}-f_1g_1\right]-\sum_{r=1}^n g_{r+1}(f_{r+1}-f_r)$$ Putting $f_r=r^2$ and $g_r=\frac {r(r-1)(2r-1)}6$ gives $$\sum_{r=1}^n r^4=(n+1)^2\cdot \frac {n(n+1)(2n+1)}6-\sum_{r=1}^n \frac {r(r+1)(2r+1)}6\cdot (2r+1)$$ but again this does not seem to get us any further.


1st Edit

Putting $T_m=\sum_{r=1}^n r^m$, the original problem can be restated as an attempt to prove that

$$5T_4=(6T_1-1)T_2$$


2nd Edit

This paper might be useful.

In section 4 (p$206$), it is stated that $$\frac{\sigma_4}{\sigma_2}=\frac {6\sigma_1-1}5$$ which is derived from the Faulhabner polynomials. $\sigma_m$ has the same definition as our $T_m$ as defined above.

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Statement we'd like to show: $$ \sum^{n}_{r=1}r^4=\frac{3n^2+3n-1}{5}\sum^{n}_{r=1}r^2 $$ Or equivalently: $$ 5\sum^{n}_{r=0}r^4=(3n^2+3n-1)\sum^{n}_{r=0}r^2 $$

We may recall one of approaches how to get closed form of this expression and utilize it in a slightly different way. $$ \sum^{n}_{r=0}r^5 + (n+1)^5=\sum^{n}_{r=0}(r+1)^5\\ (r+1)^5=r^5+5r^4+10r^3+10r^2+5r+1 $$ Which leads to: $$ (n+1)^5=5\sum^{n}_{r=0}r^4+10\sum^{n}_{r=0}r^3+10\sum^{n}_{r=0}r^2+5\sum^{n}_{r=0}r+\sum^{n}_{r=0}1 $$ Similar actions give us: $$ (n+1)^4=4\sum^{n}_{r=0}r^3+6\sum^{n}_{r=0}r^2+4\sum^{n}_{r=0}r+\sum^{n}_{r=0}1\\ (n+1)^3=3\sum^{n}_{r=0}r^2+3\sum^{n}_{r=0}r+\sum^{n}_{r=0}1 $$ So let us combine it: $$ 5\sum^{n}_{r=0}r^4=(n+1)^5-\left(10\sum^{n}_{r=0}r^3+10\sum^{n}_{r=0}r^2+5\sum^{n}_{r=0}r+\sum^{n}_{r=0}1\right)\to\\ 5\sum^{n}_{r=0}r^4=(n+1)^5-\left(\frac{5}{2}(n+1)^4-5\sum^{n}_{r=0}r^2-5\sum^{n}_{r=0}r-\frac{3}{2}\sum^{n}_{r=0}1\right)\to\\ 5\sum^{n}_{r=0}r^4=(n+1)^5-\left(\frac{5}{2}(n+1)^4-\frac{3}{2}(n+1)^3-\frac{1}{2}\sum^{n}_{r=0}r^2-\frac{1}{2}\sum^{n}_{r=0}r\right)\to\\ 5\sum^{n}_{r=0}r^4=n(n+1)^3\left(n-\frac{1}{2}\right)+\frac{1}{2}\sum^{n}_{r=0}r+\frac{1}{2}\sum^{n}_{r=0}r^2 $$ At this point we have to express $\sum^{n}_{r=0} r=\frac{n(n+1)}{2}$, which after some simplifications leads us to: $$ 5\sum^{n}_{r=0}r^4=3\left(n^2+n-\frac{1}{2}\right)\frac{n(n+1)(2n+1)}{6}+\frac{1}{2}\sum^{n}_{r=0}r^2 $$ We recognize $\sum^{n}_{r=0}r^2=\frac{n(n+1)(2n+1)}{6}$, and finish with: $$ 5\sum^{n}_{r=0}r^4=(3n^2+3n-1)\sum^{n}_{r=0}r^2 $$

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  • $\begingroup$ Thanks for your solution! (+1) $\endgroup$ – hypergeometric Oct 2 '16 at 23:30

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