An algebra problem by Akshat Sharda

write as $\dfrac{3}{n+2}\left(\dfrac{3}{(n-1)(n+2)}\right)=\dfrac{3}{n+2}\left(\dfrac{1}{n-1}-\dfrac{1}{n+2}\right)=\dfrac{3}{(n-1)(n+2)}-\dfrac{3}{(n+2)^2}$ $=\dfrac{1}{n-1}-\dfrac{1}{n+2}-\dfrac{3}{(n+2)^2}$ so the summation becomes $\sum_{n=2}^\infty \left(\dfrac{1}{n-1}\right)-\sum_{n=2}^\infty \left(\dfrac{1}{n+2}\right)-3\sum_{n=2}^\infty \left(\dfrac{1}{(n+2)^2}\right)$ first part is an easy telescoping series. the second one is the solution to the basel problem (or equivalently $\zeta(2)$ ). the summation becomes: $\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{3\pi^2}{6}+\dfrac{3}{1^2}+\dfrac{3}{2^2}+\dfrac{3}{3^2}$ this equals: $\dfrac{71-6\pi^2}{12}$ hence: $71+6+2+12=\boxed{91}$ . note that the problem doesnot specify anything about abcd. many possible answers.