Not easy as sum of squares -2

Algebra Level 5

$\sum_{ k=1 }^\infty{ \frac { 1 }{ { k }^{ 2 } } =\frac { { \pi }^{ 2 } }{ 6 } },\qquad { S }_{ i }=\sum_{ k=1 }^\infty\frac { i }{ { \big(36{ k }^{ 2 }-1\big) }^{ i } }$

Given the above, $S_{1}+S_{2}$ can be represented as $\dfrac { { \pi }^{ 2 } }{ a } -\dfrac { b }{ c }$ .