If $$\frac{1}{3^2 + 1} + \frac{1}{4^2 + 2} + \frac{1}{5^2 + 3} + \cdots = \frac{a}{b}$$, where $$a$$ and $$b$$ are coprime natural numbers, then $$a + b =$$

The general term is $$\frac{1}{n^2 + n - 2} = \frac{1}{(n - 1)(n + 2)} = \frac{1}{3}\left(\frac{1}{n - 1} - \frac{1}{n + 2}\right)$$ for $$n = 3, 4, 5, \ldots$$. The series telescopes, leaving $$\frac{1}{3}\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right) = \frac{1}{3} \times \frac{13}{12} = \frac{13}{36}$$. Since 13 and 36 are coprime, $$a + b = 13 + 36 = 49$$.