Question: $$\sum r(r+1)(r+2)(r+3)$$ is equal to?

Here, $r$ varies from $1$ to $n$

I am having difficulty in solving questions involving such telescoping series. While I am easily able to do questions where a term splits into two terms, such question are not coming to me.

I am getting no ideas on how to split such a large term. Maybe a hint or help would get me moving.

Answer: 
By telescopic sum we obtain:
$$\sum\_{r=1}^nr(r+1)(r+2)(r+3)=$$
$$=\frac{1}{5}\sum\_{r=1}^n(r(r+1)(r+2)(r+3)(r+4)-(r-1)r(r+1)(r+2)(r+3))=$$
$$=\frac{1}{5}n(n+1)(n+2)(n+3)(n+4)$$
Notice that $r(r+1)(r+2)(r+3)=24\binom{r+3}{4}$ then exploit the [Hockey-stick identity](https://en.wikipedia.org/wiki/Hockey-stick_identity).