Easier way to solve $1-\frac1{2^2}-\frac1{4^2}+\frac1{5^2}+\frac1{7^2}-\frac1{8^2}-\frac1{10^2}\ldots=\frac49s$? 
Show that $1-\frac1{2^2}-\frac1{4^2}+\frac1{5^2}+\frac1{7^2}-\frac1{8^2}-\frac1{10^2}++--\ldots=\frac49s$, where $s=\sum k^{-2}$

I can divide the LHS in the sum of four series
$$1-\frac1{2^2}-\frac1{4^2}+\frac1{5^2}+\frac1{7^2}-\frac1{8^2}-\frac1{10^2}++--\ldots=\\=\sum_{k\ge 1}\frac1{(6k-1)^2}+\sum_{k\ge 1}\frac1{(6k-5)^2}-\sum_{k\ge 1}\frac1{(6k-2)^2}-\sum_{k\ge 1}\frac1{(6k-4)^2}=\frac49\sum_{k\ge 1}\frac1{k^2}$$
but manipulating the above expression is so tedious to show the equality. Do you know a different way to solve this problem? Thank you.
P.S.: I dont know exactly what kind of tags use for this question.

I don't  know if it is an easy way. But I say it anyway.
To make the required sum, we may start from
$S=\sum\frac{1}{k^2}$
Then, we wanna take all the terms of form $\frac{1}{(3k)^2}$ off.
Then, take $2$ times the terms of the form $\frac{1}{(2k)^2}$ off.
As there is an intersection between the terms of the form $\frac{1}{(3k)^2}$ and $\frac{1}{(2k)^2}$, $2$ times the terms of the form $\frac{1}{(6k)^2}$ is added.
The final result is
$\sum\frac{1}{k^2}-\sum\frac{1}{(3k)^2}-2\sum\frac{1}{(2k)^2}+2\sum\frac{1}{(6k)^2}=S-\frac{1}{9}S-\frac{1}{2}S+\frac{1}{18}S=\frac{4}{9}S$
