Telescopic summation

$\begin{aligned} S & = \frac 1{3^2+1} + \frac 1{4^2+2} + \frac 1{5^2+3} + \frac 1{6^2+4} + ... \\ & = \sum_{k=1}^\infty \frac 1{(k+2)^2+k} \\ & = \sum_{k=1}^\infty \frac 1{k^2+4k+4+k} \\ & = \sum_{k=1}^\infty \frac 1{k^2+5k+4} \\ & = \frac 13 \sum_{k=1}^\infty \left(\frac 1{k+1} - \frac 1{k+4} \right) \\ & = \frac 13 \left(\frac 12 + \frac 13 + \frac 14 \right) \\ & = \frac {13}{36} \end{aligned}$

$\implies m + n = 13+36 = \boxed{49}$

$\sum_{n=1}^\infty \dfrac{1}{(n+2)^2 + n} = \sum_{n=1}^\infty \dfrac{1}{n^2 + 4n + 4 + n} = \sum_{n=1}^\infty \dfrac{1}{(n+1)(n+4)} \\ \sum_{n=1}^\infty \dfrac{-3 + n - n}{-3(n+1)(n+4)} = \dfrac{1}{-3}\sum_{n=1}^\infty \dfrac{(n+1) - (n+4)}{-2(n+1)(n+4)} \\ = \dfrac{1}{-3} \sum_{n=1}^\infty (\color{#D61F06}{\dfrac{1}{n+4}} - \color{#3D99F6}{\dfrac{1}{n+1})} \\ = \dfrac{1}{-3} (\color{#D61F06}{\dfrac{1}{5} + \dfrac{1}{6} + \ldots} - \color{#3D99F6}{\dfrac{1}{2} - \dfrac{1}{3} - \dfrac{1}{4} - \dfrac{1}{5} - \dfrac{1}{6} - \ldots}) \\ = \dfrac{1}{-3}(\dfrac{1}{2} - \dfrac{1}{3} - \dfrac{1}{4}) \\ = \dfrac{1}{4} (\dfrac{13}{12}) = \dfrac{13}{36} \\ \boxed{\therefore m + n = 49}$