<s>[INST] 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:
$$= \dfrac{1}{(k-1)d}\left(\frac{1}{r(r+d)...(r+(k-2)d)} - \frac{1}{(r+d)(r+2d)...(r+(k-1)d)}\right)$$
 [/INST] 
And solving them takes much time.

There is a trick so solving partial fractions quickly.
$$\frac{1}{(r)(r + d)(r + 2d)(r + 3d)}
=
\frac{A}{r} + \frac{B}{r + d} + \frac{C}{r + 2d} + \frac{D}{r + 3d}$$
$$1 
=
A(r + d)(r + 2d)(r + 3d) + 
B(r)(r + 2d)(r + 3d) +
C(r)(r + d)(r + 3d) + 
D(r)(r + d)(r + 2d)$$
The above is true for all $r$.  So plug in the roots of the original equation.  $r = 0$ gives:
$$1 = A(d)(2d)(3)$$
$$A = \frac{1}{6d^3}$$
Then $r = -d$ :
$$1 = B(-d)(d)(2d)$$
$$B = \frac{-1}{2d^3}$$
Then $r = -2d$:
$$1 = C(-2d)(-d)(d)$$
$$C = \frac{1}{2d^3}$$
Last $r = -3d$:
$$1 = D(-3d)(-2d)(-d)$$
$$D = \frac{-1}{6d^3}$$
So you get:
$$S = \frac{1}{6d^3}\left( \frac{1}{r} - \frac{3}{r + d} + \frac{3}{r + 2d} - \frac{1}{r + 3d}\right)$$
After you do it a few times and get the pattern, you can almost do it in your head.  I suggest looking at it as 2 telescoping series:
$$\text{Sum} = \left(\frac{1}{6d^3} \sum_{r = 1}^n \frac{1}{r} - \frac{1}{r + 3d}\right) + 
 \left(\frac{-1}{2d^3} \sum_{r = 1}^n \frac{1}{r + d} - \frac{1}{r + 2d}\right)$$
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