Finding sum of the series $\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$ Find the sum: $$\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$$
My method:
I tried to split it into partial fractions like: $\dfrac{A}{r}, \dfrac{B}{r+d}$ etc. Using this method, we have 4 equations in $A,B,C,D$! And solving them takes much time. I split into partial fractions hoping that some of them will cancel out and the sum might telescope.
I hope theres a simpler method to this.
Edit
As a side-question, can we extend this to $k$ factors in denominator, something like
$$\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)...(r+(k-1)d)}$$
Using N.S answer, we can extend it easily:
$$= \dfrac{1}{(k-1)d}\left(\frac{1}{r(r+d)...(r+(k-2)d)} - \frac{1}{(r+d)(r+2d)...(r+(k-1)d)}\right)$$

(Modified answer)
Define $r^{\overline{m}(d)}$ as the $d$-tuple rising factorial, i.e. 
$r^{\overline{m}(d)}=r(r+d)(r+2d)\cdots(r+(m-1)d) $ 
and $r^{-\overline{m}(d)}=\frac 1{r^{\overline{m}(d)}}$
, e.g. $r^{\overline{4}(3)}=r(r+3)(r+6)(r+9)$.
$$\begin{align}
S(m,d)&=\sum_{r=1}^n\frac 1{r(r+d)(r+2d)\cdots (r+(m-1)d)}\\
&=\sum_{r=1}^n r^{-\overline{m}(d)}\\
&=\frac 1{(m-1)d}\sum_{r=1}^n r^{-\overline{m-1}(d)}\ -\ (r+d)^{-\overline{m-1}(d)}\\
&=\boxed{\frac 1{(m-1)d}\sum_{r=1}^d  r^{-\overline{m-1}(d)}\ - (n+r)^{-\overline{m-1}(d)} }
\end{align}$$
For example, if $d=3, m=4, n=100$, 
$$\small\begin{align}
&S(3,4)\\
&=\sum_{r=1}^{100}\frac 1{r(r+3)(r+6)(r+9)}\\
&\color{lightgrey}{=\small\frac 1{1\cdot 4\cdot7\cdot 10}
+\frac 1{2\cdot5\cdot8\cdot 11}+\frac 1{3\cdot6\cdot9\cdot 12}
+\cdots+
\frac 1{100\cdot103\cdot106\cdot109}
}\\
&=\frac 19 \sum_{r=1}^{100}\frac 1{r(r+3)(r+6)}-\frac 1{(r+3)(r+6)(r+9)}\\
&=\frac 19\left[ \sum_{r=1}^{100}\frac 1{r(r+3)(r+6)}-\sum_{r=4}^{103}\frac 1{r(r+3)(r+6)}\right]\\
&=\frac 19\left[ \sum_{r=1}^3 \frac1{r(r+3)(r+6)}-\sum_{r=101}^{103}\frac 1{r(r+3)(r+6)(r+9)}\right]\\
&=\small\frac 19 \left[\frac 1{1\cdot 4\cdot7\cdot}
+\frac 1{2\cdot5\cdot8}+\frac 1{3\cdot6\cdot9}
-\frac 1{101\cdot 104\cdot 107}
-\frac 1{102\cdot 105\cdot 108}
-\frac 1{103\cdot106\cdot 109}
\right]\end{align}$$
