Three summations 
Given
  $$S_n=\sum^{n-1}_{i=0}\sum^{i-1}_{j=0}\sum^{j-1}_{k=0} (i+j+k), $$
  there are  positive integers $A$ and $B$ such that
  $$\frac{1}{S_3}+\frac{1}{S_4}+\frac{1}{S_5}+\dots=A-\frac{2\pi^2}{B}$$
  Find $A+B$.

MyApproach:
I need to solve the innermost summation first and then proceed to the last one. But I did not proceed. The summation for $N$ natural numbers are $n(n+1)/2$.
Any help?
 A: You are on the right track. Following your approach, show first that 
$$4S_n=n(n-2)(n-1)^2.$$ 
Therefore, by partial fraction decomposition,
$$\frac{1}{S_n}=\frac{2}{n-1}-\frac{2}{n}-\frac{4}{(n-1)^2}.$$
Hence, for $N\geq 3$
$$\sum_{n=3}^N\frac{1}{S_n}=2\sum_{n=3}^N \left(\frac{1}{n-1}-\frac{1}{n}\right)-4\sum_{n=3}^N \frac{1}{(n-1)^2}.$$
The first sum is telescopic and the second one is related to $\sum_{n=1}^{\infty}1/n^2=\pi^2/6$.
Can you take it from here?
A: Expanding the triple summation:
$$\begin{align} S_n=&\sum^{n-1}_{i=0}\sum^{i-1}_{j=0}\sum^{j-1}_{k=0} (i+j+k)=\\ &\sum^{n-1}_{i=0}\sum^{i-1}_{j=0} \left(ij+j^2+\frac{j(j-1)}{2}\right)=\\ &\sum^{n-1}_{i=0}\left(i\cdot \frac{i(i-1)}{2}+\frac{3}{2}\cdot \frac{(i-1)i(2i-1)}{6}-\frac{1}{2}\cdot \frac{(j-1)j}{2}\right)=\\
\frac14&\sum^{n-1}_{i=0}\left(4i^3-6i^2+2i\right)=\\
\frac14&\left(4\cdot \frac{(n-1)^2n^2}{4}-6\cdot\frac{(n-1)n(2n-1)}{6}+2\cdot\frac{(n-1)n}{2}\right)=\\
\frac14&n(n-1)^2(n-2).\end{align}$$
Hence:
$$\begin{align}\frac{1}{S_n}=\frac{4}{n(n-1)^2(n-2)}=&4\left(\frac{1}{n(n-2)}-\frac{1}{(n-1)^2}\right)=\\
&4\left(\frac{1}{2(n-2)}-\frac{1}{2n}-\frac{1}{(n-1)^2}\right)=\\ &\frac{2}{n-2}-\frac{2}{n}-\frac{4}{(n-1)^2}.\end{align}$$
So:
$$\sum_{n=3}^{\infty}\frac{1}{S_n}=2\sum_{n=3}^{\infty} \left(\frac{1}{n-2}-\frac{1}{n}\right)-4\sum_{n=3}^{\infty} \frac{1}{(n-1)^2}=\\
2\cdot \frac32-4\cdot \frac{\pi^2}{6}=3-\frac{2\pi^2}{3}.$$
Finally, $A+B=3+3=6$.
