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:
$$\sum\_{r=1}^nr(r+1)(r+2)(r+3)=$$
$$=\frac{1}{5}\sum\_{r=1}^n(r(r+1)(r+2)(r+3)(r+4)-(r-1)r(r+1)(r+2)(r+3))=$$
$$=\frac{1}{5}n(n+1)(n+2)(n+3)(n+4)$$
To generalize your question concerning 
a very important polynomial (function), note that
$$
r\left( {r + 1} \right) \cdots \left( {r + m - 1} \right) = \prod\limits\_{0\, \le \,k\, \le \,m - 1} {\left( {r + k} \right)} = r^{\,\overline {\,m\,} } = {{\Gamma (r + m)} \over {\Gamma (r)}}
$$
is called [*Rising Factorial (Pochammer symbol)*](https://en.wikipedia.org/wiki/Falling_and_rising_factorials)

Among the many properties, it has the fact that its **finite difference** resembles the derivative of $x^m$,
as the sum resembles the integral
$$
\eqalign{
 & \Delta \_{\,r} \;r^{\,\overline {\,m\,} } = \left( {r + 1} \right)^{\,\overline {\,m\,} } - r^{\,\overline {\,m\,} } = m\;\left( {r + 1} \right)^{\,\overline {\,m - 1\,} } \cr 
 & \sum {r^{\,\overline {\,m\,} } } \; = \Delta \_{\,r} ^{\left( { - 1} \right)} \;r^{\,\overline {\,m\,} } = {1 \over {m + 1}}\;\left( {r - 1} \right)^{\,\overline {\,m + 1\,} } + c \cr} 
$$
where $\sum$ is the [*Antiderivative* or *Indefinite Sum*](https://en.wikipedia.org/wiki/Indefinite_sum)

In your particular case
$$
\eqalign{
 & \sum {r\left( {r + 1} \right)\left( {r + 2} \right)\left( {r + 3} \right) = } \sum {r^{\,\overline {\,4\,} } } \; = {1 \over 5}\;\left( {r - 1} \right)^{\,\overline {\,5\,} } + c = \cr 
 & = {1 \over 5}\;\left( {r - 1} \right)r\left( {r + 1} \right)\left( {r + 2} \right)\left( {r + 3} \right) + c \cr} 
$$

and for instance:
$$
\eqalign{
 & \sum\limits\_{r = 1}^n {r\left( {r + 1} \right)\left( {r + 2} \right)\left( {r + 3} \right)} = \left. {{1 \over 5}\;\left( {r - 1} \right)^{\,\overline {\,5\,} } } \right|\_{\;1}^{\;n + 1} = {1 \over 5}\left( {\;n^{\,\overline {\,5\,} } - 0^{\,\overline {\,5\,} } } \right) = \cr 
 & = {1 \over 5}\;n^{\,\overline {\,5\,} } = {1 \over 5}\;n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {n + 4} \right) \cr} 
$$
note that the definite sum from $a$ to $b$, corresponds
to the indefinite sum calculated at **$b+1$** minus that calculated at $a$.