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

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 1

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.

Answer 2

By telescopic sum we obtain: $$\sum_{r=1}^nr(r+1)(r+2)(r+3)=$$ $$=\frac{1}{5}\sum_{r=1}^n(r(r+1)(r+2)(r+3)(r+4)-(r-1)r(r+1)(r+2)(r+3))=$$ $$=\frac{1}{5}n(n+1)(n+2)(n+3)(n+4)$$

Answer 3

Notice that $r(r+1)(r+2)(r+3)=24\binom{r+3}{4}$ then exploit the Hockey-stick identity.

Answer 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)

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

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 $a$.