So I've tried expressing this in a few different ways but online I've only found solutions in log base 10...
$\sum_{n=1}^{\infty}\frac{(\frac{1}{3})^n + (\frac{2}{3})^n}{n^2}$
The question is to express this in the form $\frac{Q\pi^W+Eln(R)ln(T)+Y(ln(U))^I}{O}$. Is there something I'm missing? Thanks
Note that for $-1\le x<1$, $\sum_{n=1}^\infty \frac{x^{n-1}}{n}=-\frac{\log(1-x)}{x}$. Integration reveals
$$\sum_{n=1}^\infty \frac{x^{n}}{n^2}=-\int_0^x \frac{\log(1-t)}t\,dt=\text{Li}_2(x)$$
where $\text{Li}_2(x)=-\int_0^x \frac{\log(1-t)}{t}\,dt$ is the dilogarithm function.
Therefore, we can write
$$\begin{align} \sum_{n=1}^\infty \frac{\left(\frac13\right)^n+\left(\frac23\right)^n}{n^2}&=-\int_0^{1/3} \frac{\log(1-t)}t\,dt-\int_0^{2/3} \frac{\log(1-t)}t\,dt\\\\ &=\text{Li}_2(1/3)+\text{Li}_2(2/3)\\\\ &=\text{Li}_2(1/3)+\text{Li}_2(1-1/3)\\\\ &=\frac{\pi^2}{6}-\log(1/3)\log(2/3) \end{align}$$
where we used the identity $\text{Li}_2(x)+\text{Li}_2(1-x)=\frac{\pi^2}{6}-\log(x)\log(1-x)$.