<s>[INST] Proving $\sum _{r=1}^n r^4$ by considering $\sum _{r=1} ^n \left( (r+1)^5 - r^5 \right)$ 
By considering $$\sum _{r=1} ^n \left( (r+1)^5 - r^5 \right)$$ Show that $$\sum _{r=1}^n r^4 = \frac 1 {30} n(n+1)(2n+1)(3n^2 + 3n-1)$$

So I'm a little unsure on how to really start on this so I've expanded the given expression an found that 
$$\sum _{r=1} ^n \left( (r+1)^5 - r^5 \right)  = \sum_{r=1} ^n 5r^4 +10r^3 + 10r^2 +5r+1$$
I know I want to get this into a form which I can then solve for $\sum _{r=1}^n r^4 $ using the fact I know the sums of $\sum _{r=1}^n r^3$ and $\sum _{r=1}^n r^2$. If anyone could helping me get started with this question that would be great.
 [/INST] OK, i will also type the "same answer", but am not that quick...

We use below a natural $n$. Let $S(n,p)$ be the sum of the numbers $1^p, 2^p,\dots, n^p$. Then:
$$
\begin{aligned}
(n+1)^5-1^5
&=\sum_{1\le r\le n}((r+1)^5-r^5)\\
&=5S(n,4)+10S(n,3)3+10S(n,2)+5S(n,1)+S(n,0)\\
&=5S(n,4)
+\frac {10}4\cdot n^2(n+1)^2
+\frac {10}6\cdot n(n+1)(2n+1)
+\frac 52\cdot n(n+1)
+n
\\[2mm]
&\qquad\text{ so}
\\[2mm]
30S(n,4)
&=
6((n+1)^5-1)
-15 n^2(n+1)^2
-10 n(n+1)(2n+1)
-15 n(n+1)
-6 n
\\
&=
6n(n^4+5n^3+10n^2+10n+5)
-15 n^2(n+1)^2
-10 n(n+1)(2n+1)
-15 n(n+1)
-6 n
\\
&=
6n(n^4+5n^3+10n^2+10n+4)
-15 n^2(n+1)^2
-10 n(n+1)(2n+1)
-15 n(n+1)
\\
&=
6n(n+1)(n^3+4n^2+6n+4)
-15 n^2(n+1)^2
-10 n(n+1)(2n+1)
-15 n(n+1)
\\
&=
n(n+1)\Big(\ 6(n^3+4n^2+6n+4)-15(n^2+n)-10(2n+1)-15\ \Big)
\\
&=
n(n+1)\Big(\ 3(n^2 + n + 3)(2n + 1)-10(2n+1)\ \Big)
\\
&=
n(n+1)(2n+1)\Big(\ 3n^2 + 3n + 9-10\ \Big)
\\
&=
n(n+1)(2n+1)\Big(\ 3n^2 + 3n + 1\ \Big)
\end{aligned}
$$
</s>