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.

Since $(1,-1,0,-1,1,0)=(1,-1,1,-1,1,-1)-(0,0,1,0,0,-1)$ we have:
$$\begin{eqnarray*} S&=&\sum_{k\geq 0}\left(\frac{1}{(6k+1)^2}-\frac{1}{(6k+2)^2}-\frac{1}{(6k+4)^2}+\frac{1}{(6k+5)^2}\right)\\&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}-\sum_{n\geq 1}\frac{(-1)^{n+1}}{(3n)^2}\\&=&\frac{8}{9}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}\end{eqnarray*}$$
and
$$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2} = \sum_{n\geq 1}\frac{1}{n^2}-2\sum_{n\geq 1}\frac{1}{(2n)^2} = \frac{1}{2}\sum_{n\geq 1}\frac{1}{n^2}$$
hence $S=\frac{4}{9}\zeta(2)=\color{red}{\large\frac{2\pi^2}{27}}$ as wanted.
