# Difference between revisions of "2020 AMC 12A Problems/Problem 22"

## Problem

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$$\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?$$ $\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47$

## Solution 1

Square the given equality to yield $$(3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,$$ so $a_nb_n = \tfrac12\operatorname{Im}((3+4i)^n)$ and $$\sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) = \boxed{\frac 7{16}}.$$

## Solution 2 (DeMoivre's Formula)

We rewrite $$(2+i)^n=\left(\sqrt{5}\left(\frac{2}{\sqrt{5}}+\frac{i}{\sqrt{5}}\right)\right)^n=5^{\frac{n}{2}}e^{in\tan^{-1}\left(\frac{1}{2}\right)}=5^{\frac{n}{2}}\left(\cos \left(n\tan^{-1}\left(\frac{1}{2}\right)\right)+i\sin\left(n\tan^{-1}\left(\frac{1}{2}\right)\right)\right)$$ by DeMoivre's Formula. Letting $\theta=\tan^{-1}\left(\frac{1}{2}\right)$, we know that $a_n=5^{\frac{n}{2}}\cos \left(n\theta\right)$ and $b_n=5^{\frac{n}{2}}\sin \left(n\theta\right)$. The desired sum then turns into $$\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^n\cos\left(n\theta\right)\sin\left(n\theta\right)$$ $$=\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^n\sin\left(2n\theta\right)=\frac{1}{2}\text{Im}\left(\sum_{n=0}^{\infty}\left(\frac{5}{7}\right)^ne^{2in\theta}\right)$$ This is now an infinite geometric series! After finding $\sin(2\theta)=2\cdot \frac{2}{\sqrt{5}}\cdot \frac{1}{\sqrt{5}}=\frac{4}{5}$, $\cos(2\theta)=\sqrt{1-\sin^2(2\theta)}=\frac{3}{5}$, we find $$S=\frac{1}{2}\text{Im}\left(\frac{1}{1-\frac{5}{7}\cdot e^{2i\theta}}\right)=\frac{1}{2}\text{Im}\left(\frac{1}{1-\frac{3}{7}-\frac{4}{7}i}\right)=\frac{1}{2}\cdot \frac{\frac{4}{7}}{\frac{32}{49}}=\boxed{\textbf{(B) }\frac{7}{16}}$$ ~ktong