A 2nd attempt

Firstly, the sum can be rewritten in a better form as follows: $\displaystyle \sum_{i=0}^{\infty} \frac{18i^2 +2}{(9i^2 - 1)^2} = \displaystyle \sum_{i=0}^{\infty} \frac{(3i-1)^2 + (3i+1)^2}{(3i-1)^2 (3i+1)^2} =\displaystyle \sum_{i=0}^{\infty} \frac{1}{(3i-1)^2} + \frac{1}{(3i+1)^2}$ Now $\displaystyle \sum_{i=0}^{\infty} \frac{1}{(3i+1)^2} = \displaystyle \sum_{i=1}^{\infty} \frac{1}{i^2} - \displaystyle \sum_{i=1}^{\infty} \frac{1}{(3i)^2} - \displaystyle \sum_{i=1}^{\infty} \frac{1}{(3i-1)^2}$ $=\frac{8}{9} \displaystyle \sum_{i=1}^{\infty} \frac{1}{i^2} - \displaystyle \sum_{i=1}^{\infty} \frac{1}{(3i-1)^2} =\frac{4\pi^2}{27} - \displaystyle \sum_{i=1}^{\infty} \frac{1}{(3i-1)^2}$ Now it is important to notice that $\displaystyle \sum_{i=1}^{\infty} \frac{1}{(3i-1)^2} + 1= \displaystyle \sum_{i=0}^{\infty} \frac{1}{(3i-1)^2}$ $\therefore\displaystyle \sum_{i=0}^{\infty} \frac{1}{(3i-1)^2} + \frac{1}{(3i+1)^2} = \frac{4\pi^2}{27} + 1 = 2.46 \ (3s.f)$