Help to compute sum of products

HINT:

$$(r-1)r(r+1)=r^3-r$$

Now, $$\sum_{r=1}^n r=\frac{n(n+1)}2$$ and $$\sum_{r=1}^n r^3=\left(\frac{n(n+1)}2\right)^2$$

Here $r=1$ to $21$

Hint: Note that $(n+1)(n+2)(n+3)(n+4)-(n)(n+1)(n+2)(n+3)=4(n+1)(n+2)(n+3)$.

Using this identity write our sum as a collapsing (telescoping) sum. It may help to look at $4$ times our sum.

There is a very short solution, using difference calculus, which is a theory underlying Andre Nicolas' hint.

$$\sum_{k=0}^{23}k^{\underline{3}}\delta k=\frac{1}{4}k^{\underline{4}}|_0^{23}=\frac{1}{4}(23^{\underline{4}}-0^{\underline{4}})=\frac{23\cdot 22\cdot 21\cdot 20}{4}=53,130$$