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.
 A: 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.
A: 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
$\sum\frac{1}{k^2}-\sum\frac{1}{(3k)^2}-2\sum\frac{1}{(2k)^2}+2\sum\frac{1}{(6k)^2}=S-\frac{1}{9}S-\frac{1}{2}S+\frac{1}{18}S=\frac{4}{9}S$
A: If you are familiar with Euler products, you can see that
$$1+{1\over5^2}+{1\over7^2}+{1\over11^2}+\cdots=\prod_{p\not=2,3}\left(1-{1\over p^2}\right)^{-1}=\left(1-{1\over 2^2}\right)\left(1-{1\over 3^2}\right)\prod_p\left(1-{1\over p^2}\right)^{-1}\\={3\over4}\cdot{8\over9}\left(1+{1\over2^2}+{1\over3^3}+{1\over4^2}+\cdots\right)={2\over3}s$$
and
$${1\over2^2}+{1\over4^2}+{1\over8^2}+{1\over10^2}+\cdots={1\over2^2}\left(1+{1\over2^2}+{1\over4^2}+{1\over5^2}+\cdots\right)={1\over4}\prod_{p\not=3}\left(1-{1\over p^2}\right)^{-1}={1\over4}\left(1-{1\over 3^2}\right)\prod_p\left(1-{1\over p^2}\right)^{-1}={1\over4}\cdot{8\over9}\left(1+{1\over2^2}+{1\over3^3}+{1\over4^2}+\cdots\right)={2\over9}s$$
It follows that
$$1-{1\over2^2}-{1\over4^2}+{1\over5^2}+{1\over7^2}-{1\over8^2}-{1\over10^2}++--\cdots={2\over3}s-{2\over9}s={4\over9}s$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{f}\pars{\mu} \equiv
\sum_{k = 1}^{\infty}{1 \over \pars{6k - \mu}^{2}}\,;\qquad
\mrm{f}\pars{1} - \mrm{f}\pars{2} - \mrm{f}\pars{4} + \mrm{f}\pars{5}:\ ?}$.

\begin{align}
\mrm{f}\pars{\mu} & \equiv
\sum_{k = 1}^{\infty}{1 \over \pars{6k - \mu}^{2}} =
{1 \over 36}\sum_{k = 0}^{\infty}{1 \over \pars{k + 1 - \mu/6}^{2}}
\\[5mm] & =
{1 \over 36}\,\Psi\, '\pars{1 - {\mu \over 6}}\qquad\qquad\qquad\qquad\qquad
\pars{\Psi:\ Digamma\ Function}
\end{align}

\begin{align}
&\mrm{f}\pars{1} - \mrm{f}\pars{2} - \mrm{f}\pars{4} + \mrm{f}\pars{5} =
{1 \over 36}\bracks{\Psi\, '\pars{5 \over 6} - \Psi\, '\pars{2 \over 3} -
\Psi\, '\pars{1 \over 3} + \Psi\, '\pars{1 \over 6}}
\\[5mm] = &
{1 \over 36}\braces{\bracks{\Psi\, '\pars{5 \over 6} + \Psi\, '\pars{1 \over 6}} -
\bracks{\Psi\, '\pars{2 \over 3} + \Psi\, '\pars{1 \over 3}}}
\end{align}

With the Euler Reflection Formula:
\begin{align}
&\mrm{f}\pars{1} - \mrm{f}\pars{2} - \mrm{f}\pars{4} + \mrm{f}\pars{5} =
{1 \over 36}\bracks{\pi^{2}\csc^{2}\pars{\pi \over 6} -
\pi^{2}\csc^{2}\pars{\pi \over 3}}
\\[5mm] = &\
\bbx{\ds{{2 \over 27}\,\pi^{2}}} =
{4 \over 9}\pars{\pi^{2} \over 6} =
\bbox[10px,#ffe,border:1px groove navy]{\ds{{4 \over 9}\sum_{k = 1}^{\infty}{1 \over k^{2}}}} 
\end{align}
