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:
$$= \dfrac{1}{(k-1)d}\left(\frac{1}{r(r+d)...(r+(k-2)d)} - \frac{1}{(r+d)(r+2d)...(r+(k-1)d)}\right)$$

HINT: $$\frac{1}{r(r+d)(r+2d)(r+3d)} = \frac{1}{2d^2}\left(\frac{1}{r(r+3d)}-\frac{1}{(r+d)(r+2d)}\right) = \frac{1}{6d^3}\left(\frac{1}{r}-\frac{1}{r+3d}\right)-\frac{1}{2d^3}\left(\frac{1}{r+d}-\frac{1}{r+2d}\right).$$
