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:
$$\sum\_{r=1}^nr(r+1)(r+2)(r+3)=$$
$$=\frac{1}{5}\sum\_{r=1}^n(r(r+1)(r+2)(r+3)(r+4)-(r-1)r(r+1)(r+2)(r+3))=$$
$$=\frac{1}{5}n(n+1)(n+2)(n+3)(n+4)$$
Note that
$$
\sum\_{r=1}^n r(r+1)(r+2)(r+3)
= \sum\_{r=1}^n 4! \binom{r+3}{4}
= \sum\_{m=4}^{n+3} 4! \binom{m}{4}
$$
Now use Pascal's triangle.

The answer is

> 
>  $\displaystyle4! \binom{n+4}{5} = \frac15 n (n + 1) (n + 2) (n + 3) (n + 4)$
> 
> 
> 

Since you mention telescoping, note that:
$$
r(r+1)(r+2)(r+3)
= 4! \binom{r+3}{4}
= 4! \left(\binom{r+4}{5} - \binom{r+3}{5}\right)
$$
by Pascal's relation.