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:
$$= \dfrac{1}{(k-1)d}\left(\frac{1}{r(r+d)...(r+(k-2)d)} - \frac{1}{(r+d)(r+2d)...(r+(k-1)d)}\right)$$
 A: 
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:
$$\text{Sum} = \left(\frac{1}{6d^3} \sum_{r = 1}^n \frac{1}{r} - \frac{1}{r + 3d}\right) + 
 \left(\frac{-1}{2d^3} \sum_{r = 1}^n \frac{1}{r + d} - \frac{1}{r + 2d}\right)$$
A: Hint
$$\frac{1}{(r)(r+d)(r+2d)(r+3d)}=\frac{1}{3d}\left(\frac{1}{(r)(r+d)(r+2d)}-\frac{1}{(r+d)(r+2d)(r+3d)}\right)$$
A: When computing a partial fraction decomposition one can always invoke residues. In our case
$$ f(x)=\frac{1}{x(x+d)(x+2d)(x+3d)} = \frac{A_0}{x}+\frac{A_1}{x+d}+\frac{A_2}{x+2d}+\frac{A_3}{x+3d} $$
and 
$$ A_k = \text{Res}\left(f(x),x=-kd\right) = \lim_{x\to -kd}(x+kd)\,f(x)=\frac{1}{\prod_{\substack{0\leq j\leq 3 \\ j\neq k}}(-kd+jd)}$$
so $(A_0,A_1,A_2,A_3)$ is proportional to $\left(+\binom{3}{0},-\binom{3}{1},+\binom{3}{2},-\binom{3}{3}\right)$.
This approach has a great generality and it clearly shows that the coefficients appearing in the partial fraction decomposition of $\frac{1}{x(x+d)\cdots(x+Md)}$ are proportional to the binomial coefficients $\binom{M}{k}$ took with alternating signs.
A: HINT: $$\frac{1}{r(r+d)(r+2d)(r+3d)} = \frac{1}{2d^2}\left(\frac{1}{r(r+3d)}-\frac{1}{(r+d)(r+2d)}\right) = \frac{1}{6d^3}\left(\frac{1}{r}-\frac{1}{r+3d}\right)-\frac{1}{2d^3}\left(\frac{1}{r+d}-\frac{1}{r+2d}\right).$$
A: (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 $d=3, m=4, n=100$, 
$$\small\begin{align}
&S(3,4)\\
&=\sum_{r=1}^{100}\frac 1{r(r+3)(r+6)(r+9)}\\
&\color{lightgrey}{=\small\frac 1{1\cdot 4\cdot7\cdot 10}
+\frac 1{2\cdot5\cdot8\cdot 11}+\frac 1{3\cdot6\cdot9\cdot 12}
+\cdots+
\frac 1{100\cdot103\cdot106\cdot109}
}\\
&=\frac 19 \sum_{r=1}^{100}\frac 1{r(r+3)(r+6)}-\frac 1{(r+3)(r+6)(r+9)}\\
&=\frac 19\left[ \sum_{r=1}^{100}\frac 1{r(r+3)(r+6)}-\sum_{r=4}^{103}\frac 1{r(r+3)(r+6)}\right]\\
&=\frac 19\left[ \sum_{r=1}^3 \frac1{r(r+3)(r+6)}-\sum_{r=101}^{103}\frac 1{r(r+3)(r+6)(r+9)}\right]\\
&=\small\frac 19 \left[\frac 1{1\cdot 4\cdot7\cdot}
+\frac 1{2\cdot5\cdot8}+\frac 1{3\cdot6\cdot9}
-\frac 1{101\cdot 104\cdot 107}
-\frac 1{102\cdot 105\cdot 108}
-\frac 1{103\cdot106\cdot 109}
\right]\end{align}$$
