Need help with De'Moivre's Theorem, or maybe that is the wrong approach to this problem. 
Let ($a_{n}$) and ($b_{n}$) be the sequences of real numbers such that $(2+i)^n=a_{n}+b_{n}i$ for all integers $n\geq 0$, where $i=\sqrt{-1}$. What is $\sum_{n=0}^{\infty}\frac{a_{n}b_{n}}{7^n}$?

I need some help with the question above. I tried using De'Moivre's Theorem to simplify $(2+i)^n$, but got hung up with simplifying $\sqrt{5}^n\operatorname{cis}(n\theta)$. I was attempting to find $a_{n}$ and $b_{n}$ first, then evaluate the summation.

Let $\theta$ be the acute angle satisfying $\tan \theta = \frac{1}{2}$,
$$a_n = \sqrt{5^n} \cos(n \theta)$$
$$b_n = \sqrt{5^n}\sin (n\theta)$$
$$\frac{a_nb_n}{7^n} = \left( \frac57\right)^n\sin(n\theta)\cos(n\theta)=\frac12 \left( \frac57\right)^n \sin (2n\theta)= \frac12 \Im\left[\left(\frac57e^{2i\theta}\right)^n\right]$$
\begin{align}\sum_{n=0}^\infty \frac{a_nb_n}{7^n} &= \frac12 \Im \left[ \sum_{n=0}^\infty \left( \frac57 e^{2i\theta} \right)^n\right] \\
&= \frac12 \Im \left[ \frac{1}{1-\frac57e^{2i\theta}}\right]\end{align}
Try to simplify from here.

Edit:
\begin{align}
\frac12 \Im \left[ \frac1{1-\frac57 e^{2i\theta}}\right] &= \frac12 \Im \left[ \frac{1-\frac57e^{-2i\theta}}{(1-\frac57 e^{2i\theta})(1-\frac57 e^{-2i\theta})}\right] \\
&=\frac12\left[ \frac{\frac57\sin 2 \theta}{1+\frac{25}{49}-\frac{10}7 \cos2\theta}\right] \\
&=\frac12\left[ \frac{\frac{10}7\sin  \theta \cos \theta}{1+\frac{25}{49}-\frac{10}7 (2\cos^2 \theta -1)}\right] \\
&=\frac12\left[ \frac{\frac{10}7\cdot \frac25}{1+\frac{25}{49}-\frac{10}7 (2\cdot \frac45 -1)}\right] \\
&=\frac12\left[ \frac{\frac{10}7\cdot \frac25}{1+\frac{25}{49}-\frac{10}7 \cdot \frac35}\right] \\
&=\frac12\left[ \frac{\frac47}{1+\frac{25}{49}-\frac{6}7}\right] \\
&=\frac12\left[ \frac{28}{49+25-42}\right] \\
&= \frac{14}{32}\\
&= \frac{7}{16}
\end{align}
