How to take every third element in a series? $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = 1.644934$  or  $\frac{\pi^2}{6}$
What if we take every 3rd term and add them up? 
A = $ \frac{1}{3^2} + \frac{1}{6^2} + \frac{1}{9^2} + \cdots = ??$
How to take every 3rd-1 term and add them up?
B = $ \frac{1}{2^2} + \frac{1}{5^2} + \frac{1}{8^2} + \cdots = ??$
How to take every 3rd-2 term and add them up?
C = $ \frac{1}{1^2} + \frac{1}{4^2} + \frac{1}{7^2} + \cdots = ??$
I am not sure how to adapt Eulers methods as he used the power series of sin for his arguments: https://en.wikipedia.org/wiki/Basel_problem
 A: Note that we have
$$\psi'(z)=\sum_{n=0}^\infty \frac{1}{(n+z)^2}$$
where $\psi'(z)$ is the derivative of the digamma function.  Hence, we can write
$$\sum_{n=0}^\infty \frac{1}{(3n+1)^2}=\frac19 \psi'(1/3)$$
and 
$$\sum_{n=0}^\infty \frac{1}{(3n+2)^2}=\frac19 \psi'(2/3)$$
Interestingly, since we have
$$\sum_{n=0}^\infty \left(\frac1{(3n+3)^2}+\frac1{(3n+2)^2}+\frac1{(3n+1)^2}\right)=\frac{\pi^2}{6}$$
we find that
$$\psi'(1/3)+\psi'(2/3) = 4\pi^2/3$$
A: As a complement to Mark's answer, 
$$\sum_{n\geq 0}\frac{1}{(3n+1)^2}=-\int_{0}^{1}\sum_{n\geq 0} x^{3n}\log(x)\,dx=\int_{0}^{1}\frac{-\log x}{1-x^3}\,dx $$
(and similarly $\sum_{n\geq 0}\frac{1}{(3n+2)^2}$) can be expressed in terms of dilogarithms, since
$$ \int_{0}^{1}\frac{-\log x}{1-a x}=\frac{\text{Li}_2(a)}{a} $$
for any $|a|\leq 1$, with $\text{Li}_2(a)=\sum_{n\geq 1}\frac{a^n}{n^2}$. This is equivalent to stating that $\psi'\left(\frac{1}{3}\right)$ and $\psi'\left(\frac{2}{3}\right)$ can be computed through the discrete Fourier transform. It is worth noticing that
$$\text{Re}\,\text{Li}_2(e^{i\theta})=\sum_{n\geq 1}\frac{\cos(n\theta)}{n^2} $$
is a continuous and piecewise-parabolic function, as the formal primitive of the sawtooth wave. On the contrary, $\text{Im}\,\text{Li}_2(e^{i\theta})$ does not have a nice closed form, in general. Ref.: https://en.wikipedia.org/wiki/Spence%27s_function
A: KM101 deleted his hint... not sure why.
$\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{9^2}+\dots=\frac{1}{3^2 1^2}+\frac{1}{3^2 2^2}+\frac{1}{3^2 3^3}+\dots=\frac{1}{9}(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots)= \frac{\pi^2}{54}$
$\frac{1}{2^2}+\frac{1}{5^2}+\frac{1}{8^2}+\dots=\frac{1}{2^2 1^2}+\frac{1}{3^2?? 1^2}+\frac{1}{3^2 ??1^2}$
