Three Sum

$\large \displaystyle S_n = \sum_{i=0}^{n-1} \sum_{j=0}^{i-1} \sum_{k=0}^{j-1} (i + j + k)$

Since $\color{#20A900} \sum_{i=0}^n i = \frac{n(n+1)}{2}$ :

$\large \displaystyle S_n = \sum_{i=0}^{n-1} \sum_{j=0}^{i-1} ij + j^2 + \frac{(j-1)j}{2}$

$\large \displaystyle S_n = \sum_{i=0}^{n-1} \sum_{j=0}^{i-1} ij + \frac{3j^2}{2} - \frac{j}{2}$

Since $\color{#20A900} \sum_{i=0}^n i^2 = \frac{n(n+1)(2n+1)}{6}$ :

$\large \displaystyle S_n = \sum_{i=0}^{n-1} i \left [ \frac{(i-1)i}{2} \right ] + \frac{3}{2} \left [ \frac{(i-1)i(2i-1)}{6} \right ] - \frac{(i-1)i}{4}$

$\large \displaystyle S_n = \sum_{i=0}^{n-1} i^3 - \frac{3i^2}{2} + \frac{i}{2}$

Since $\color{#20A900} \sum_{i=0}^n i^3 = \left [ \frac{(n+1)n}{2} \right ]^2$ :

$\large \displaystyle S_n = \frac{(n-1)^2 n^2 - (n-1)n(2n-1) + (n-1)n}{4}$

$\large \displaystyle S_n = \frac{(n-1)n [ (n-1)n - (2n-1) + 1 ]}{4}$

$\large \displaystyle S_n = \frac{(n-1)n [ (n^2 - 3n + 2) ]}{4}$

$\color{#20A900} \boxed{ \large \displaystyle S_n = \frac{n(n-2)(n-1)^2}{4} }$

We're looking for:

$\large \displaystyle S = \sum_{n=3}^{\infty} \frac{1}{S_n}$

$\large \displaystyle S = 4 \sum_{n=3}^{\infty} \frac{1}{n(n-2)(n-1)^2}$

By partial fractions:

$\large \displaystyle S = 4 \left [ \sum_{n=3}^{\infty} \frac12 \left (\frac{1}{n-2} - \frac{1}{n} \right ) - \sum_{n=3}^{\infty} \frac{1}{(n-1)^2} \right ]$

The first sum will telescope, and only the first two terms of the first fraction will remain. On the second sum, we make $m = n-1$ :

$\large \displaystyle S = 4 \left [ \frac12 \left (1 + \frac12 \right ) - \sum_{m=2}^{\infty} \frac{1}{m^2} \right ]$

$\large \displaystyle S = 4 \left [\frac34 - \left ( \sum_{m=1}^{\infty} \frac{1}{m^2} - 1 \right ) \right ]$

$\large \displaystyle S = 4 \left [ \frac74 - \frac{\pi ^2}{6} \right ]$

$\color{#20A900} \boxed{ \large \displaystyle S = 7 - \frac{2\pi ^2}{3} }$

Thus:

$\color{#3D99F6} \large \displaystyle A = 7, B = 3, \boxed{\large \displaystyle A+B=10}$