Sum of the Given Series (5)

We have $\begin{aligned} P & =\dfrac{1}{n^2(n+1)^2(n+2)^2} = \left(\dfrac{1}{n(n+1)(n+2)}\right)^2 \\ P & = \left(\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\dfrac{1}{n+2}\right)^2 =\left(\dfrac{1}{n(n+2)} -\dfrac{1}{(n+1)(n+2)}\right)^2 \\ P& = \left(\dfrac{1}{2n} -\dfrac{1}{2(n+2)} -\dfrac{1}{n+1} +\dfrac{1}{n+2}\right)^2 =\left(\dfrac{1}{2n} -\dfrac{1}{n+1} +\dfrac{1}{2(n+2)}\right)^2 \end{aligned}$ Now solving algebrically we find $\begin{aligned} P & = \dfrac{1}{4n^2} -\left(\dfrac{2}{2n(n+1)}\right)+\dfrac{1}{(n+1)^2}+2\left(\dfrac{1}{2n} -\dfrac{1}{n+1}\right)\dfrac{1}{2(n+2)}+\dfrac{1}{4(n+2)^2} \\ P& =\left( \dfrac{1}{4n^2} +\dfrac{1}{(n+1)^2} +\dfrac{1}{4(n+2)^2}\right)-\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)+\dfrac{1}{4}\left(\dfrac{1}{n} -\dfrac{1}{n+2}\right)-\left(\dfrac{1}{n+1} -\dfrac{1}{n+2}\right) \\ P& = \displaystyle\sum_{n=1}^{\infty}\left[\left({\color{#3D99F6}\dfrac{1}{4n^2}+\dfrac{1}{(n+1)^2} +\dfrac{1}{4(n+2)^2}}\right) -\dfrac{3}{4n} +\dfrac{3}{4(n+2)} \right]\\ P & = {\color{#3D99F6}\dfrac{\pi^2}{4.6} +\dfrac{\pi^2}{6}-1+\dfrac{1}{4}\left(\dfrac{\pi^2}{6} -1-\dfrac{1}{4}\right)} -\dfrac{9}{8} \\ P & = \boxed{\dfrac{\pi^2}{4} -\dfrac{39}{16}} \end{aligned}$

Hence , $A+B+C+D =\boxed{61}$ .