Question

The value of $12\displaystyle \sum _{r = 1}^{\infty }\frac{\left(12 r^{2} + 1\right)}{\left(64 r^{6} - 48 r^{4} + 12 r^{2} - 1\right)}$ is :

Answer 1

$6\displaystyle \sum _{r = 1}^{10}\frac{24 r^{2} + 2}{ 64 r^{6} - 48 r^{4} + 12 r^{2} - 1}$
$=6\displaystyle \sum _{r = 1}^{\infty }\frac{24 r^{2} + 2}{\left(4 r^{2} - 1\right)^{3}}=6\displaystyle \sum T_{r}$
$\Rightarrow T_{r}=\frac{24 r^{2} + 2}{\left(2 r - 1\right)^{3} \left(2 r + 1\right)^{3}}$
$=\frac{1}{\left(2 r - 1\right)^{3}}-\frac{1}{\left(2 r + 1\right)^{3}}$
$\displaystyle \sum _{r = 1}^{\infty }T_{r}=1$