Let $(a_n)$ and $(b_n)$ be the sequences of real numbers such that\[ (2 + i)^n = a_n + b_ni \]for all integers $n\geq 0$, where $i = \sqrt{-1}$. What is\[\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?\]
We want to produce terms with $a_nb_n$. For this, let's square both sides of the given relation: $$\left((2+i)^n\right)^2=(a_n+b_ni)^2\implies (3+4i)^n = (a_n^2-b_n^2)+2a_nb_ni$$
Therefore, we find $$a_nb_n = \frac{1}{2}\Im\left((3+4i)^n\right)$$
It follows that $$\sum_{n=0}^{\infty}\frac{a_nb_n}{7^n}=\frac{1}{2}\Im\left(\sum_{n=0}^{\infty}\left(\frac{(3+4i)^n}{7^n}\right)\right)=\frac{1}{2}\Im\left(\frac{1}{1-\frac{3+4i}{7}}\right)$$
The last step utilizes the sum of infinite geometric sequence formula. The final result can be obtained as $$\frac{1}{2}\Im\left(\frac{1}{1-\frac{3+4i}{7}}\right)=\frac{7}{16}\Im(1+i)=\boxed{\frac{7}{16}}$$