Squared Triangle Sum

Since the sum of first $n$ positive integers is given by $\dfrac {n(n+1)}2$ , the given sum can be rewritten as:

$\begin{aligned} S & = \sum_{n=1}^\infty \frac 1{\left(\frac {n(n+1)}2\right)^2} = \sum_{n=1}^\infty \frac 4{n^2(n+1)^2} & \small \blue{\text{By partial fraction decomposition}} \\ & = \sum_{n=1}^\infty \left(\frac 4{(n+1)^2} + \frac 4{n^2} + \frac 8{n+1} - \frac 8n \right) & \small \blue{\text{As Riemann zeta function }\zeta(s) = \sum_{k=1}^\infty \frac 1{k^s}} \\ & = 8 \zeta(2) - 4 - 8 & \small \blue{\text{and }\zeta(2) = \frac {\pi^2}6} \\ & = \boxed{\frac {4\pi^2}3 - 12} \end{aligned}$

The sum can be expressed into the summation:

$\quad\displaystyle\sum_{n=1}^\infty \dfrac{1}{(1+2+\cdots+n)^2} \\=\displaystyle\sum_{n=1}^\infty \dfrac{1}{\left[\frac{n(n+1)}{2}\right]^2} \\= \displaystyle\sum_{n=1}^\infty \dfrac{4}{n^2(n+1)^2} \\= 4\left(\dfrac{\pi^2}{3}-3\right)=\dfrac{4\pi^2}{3}-12$

I use my question as my reference.