Sum of the first n terms of series $\frac{x^{3n}}{3n(3n-1)(3n-2)}$ I need to find sum of first n terms of series $\sum_{1}^{\infty} {\frac{x^{3n}}{3n(3n-1)(3n-2)}}$. I tried but I just don't know how to transform it into any known form of power series.
EDIT: I tried partial fraction decomposition and telescoping, something cancels but not for every $n$.
 A: $$ \frac{1}{3n(3n-1)(3n-2)} = \frac{1}{2}\left[\frac{1}{3n-2}-\frac{2}{3n-1}+\frac{1}{3n}\right] = \frac{1}{2}\int_{0}^{1}\left(z^{3n-3}-2 z^{3n-2}+z^{3n-1}\right)\,dz $$
hence
$$ \sum_{n\geq 1}\frac{x^{3n}}{3n(3n-1)(3n-2)} = \frac{1}{2}\int_{0}^{1}\sum_{n\geq 1}x^{3n} z^{3n-3}(1-z)^2\,dz=\frac{1}{2}\int_{0}^{1}\frac{x^3(1-z)^2}{1-x^3 z^3}\,dz. $$
On the other hand $\sum_{n\geq 1}\frac{x^n}{n}=-\log(1-x)$, hence
$$ \sum_{n\geq 1}\frac{x^{3n}}{3n} = -\frac{1}{3}\log(1-x^3) $$
and by letting $\omega=\exp\left(\frac{2\pi i}{3}\right)$,
$$ \sum_{n\equiv 1\!\!\pmod{3}}\frac{x^n}{n}=-\frac{1}{3}\left[\log(1-x)+\omega^2 \log(1-\omega x)+\omega \log(1-\omega^2 x)\right]$$
$$ \sum_{n\equiv 2\!\!\pmod{3}}\frac{x^n}{n}=-\frac{1}{3}\left[\log(1-x)+\omega \log(1-\omega x)+\omega^2 \log(1-\omega^2 x)\right]$$
due to the discrete Fourier transform. In particular
$$ \sum_{n\geq 1}\frac{x^{3n}}{3n(3n-1)(3n-2)}=-\frac{1}{6}\left[(x^2-2x+1)\log(1-x)+(\omega^2 x^2-2\omega x)\log(1-\omega x)+(\omega x^2-2\omega^2 x) \log(1-\omega^2 x)\right]. $$
