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.
 A: Define $S_k:=\sum_{r=1}^n n^k$ so you've shown $(n+1)^5-1=5S_4+10S_3+10S_2+5S_1+S_0$. This allows us to express $S_4$ in terms of the $S_i$ with $0\le i<4$. You could of course have used a similar approach to derives relations between earlier $S_i$. From these we can gradually prove$$S_0=n,\,S_1=\frac12n(n+1),\,S_2=\frac{1}{6}n(n+1)(2n+1),\,S_3=\frac{1}{4}n^2(n+1)^2.$$Hence $$5S_4=n\left(n^4+5n^3+10n^2+10n+5-\frac{5}{2}n(n+1)^2-\frac{5}{3}(n+1)(2n+1)-\frac{5}{2}(n+1)-1\right).$$After that it's just tedious algebra.
A: HINT:


*

*$$\small\sum _{r=1} ^n \left( (r+1)^5 - r^5 \right)=(2^5-1^5)+(3^5-2^5)+\cdots+(n^5+(n-1)^5)+((n+1)^5-n^5)=\,?$$

*$$\sum_{r=1}^n r^3=\left(\sum_{r=1}^n r\right)^2,\quad \sum_{r=1}^n r^2=\frac{n(n+1)(2n+1)}6,\quad \sum_{r=1}^n r=\frac{n(n+1)}2$$
A: 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:
$$
\begin{aligned}
(n+1)^5-1^5
&=\sum_{1\le r\le n}((r+1)^5-r^5)\\
&=5S(n,4)+10S(n,3)3+10S(n,2)+5S(n,1)+S(n,0)\\
&=5S(n,4)
+\frac {10}4\cdot n^2(n+1)^2
+\frac {10}6\cdot n(n+1)(2n+1)
+\frac 52\cdot n(n+1)
+n
\\[2mm]
&\qquad\text{ so}
\\[2mm]
30S(n,4)
&=
6((n+1)^5-1)
-15 n^2(n+1)^2
-10 n(n+1)(2n+1)
-15 n(n+1)
-6 n
\\
&=
6n(n^4+5n^3+10n^2+10n+5)
-15 n^2(n+1)^2
-10 n(n+1)(2n+1)
-15 n(n+1)
-6 n
\\
&=
6n(n^4+5n^3+10n^2+10n+4)
-15 n^2(n+1)^2
-10 n(n+1)(2n+1)
-15 n(n+1)
\\
&=
6n(n+1)(n^3+4n^2+6n+4)
-15 n^2(n+1)^2
-10 n(n+1)(2n+1)
-15 n(n+1)
\\
&=
n(n+1)\Big(\ 6(n^3+4n^2+6n+4)-15(n^2+n)-10(2n+1)-15\ \Big)
\\
&=
n(n+1)\Big(\ 3(n^2 + n + 3)(2n + 1)-10(2n+1)\ \Big)
\\
&=
n(n+1)(2n+1)\Big(\ 3n^2 + 3n + 9-10\ \Big)
\\
&=
n(n+1)(2n+1)\Big(\ 3n^2 + 3n + 1\ \Big)
\end{aligned}
$$
A: You're on the right track. Write these relations for $r=n, n-1,\dots ,2,1$ and add them. You'll have a telescoping sum  on the l.h.s. and ultimately, denoting $S_k=\sum_{r=1}^n r^k $,  you'll obtain a recursive relation between $S_4, S_3, S_2$ and $S_1$. The last three are supposed to be known, so you'll be able to deduce $S_4$.
A: You can split up sums:
$$\sum(a+b)=\sum a+\sum b$$
So use that:$$\sum(r^4)=\frac 15\bigg[\sum((r+1)^5-r^5)-10\sum(r^3)-10\sum(r^2)-5\sum r-1\bigg]$$
