Question

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 $n \ge 0,$ where $i = \sqrt{-1}.$ What is $\sum_{n=0}^{\infty} \frac{a_n b_n}{7^n}?$

$\dfrac{3}{8}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

$\dfrac{9}{16}$

$\dfrac{4}{7}$

Solution:

Since $(a_n + b_n i)^2 = a_n^2 - b_n^2 + 2 a_n b_n i,$ we have $a_n b_n = \tfrac12 \operatorname{Im}\big((2 + i)^{2n}\big) = \tfrac12 \operatorname{Im}\big((3 + 4i)^n\big).$

Therefore the sum is $\tfrac12 \operatorname{Im} \displaystyle\sum_{n=0}^{\infty} \left(\frac{3 + 4i}{7}\right)^n = \tfrac12 \operatorname{Im}\!\left(\frac{1}{1 - \frac{3 + 4i}{7}}\right).$

This equals $\tfrac12 \operatorname{Im}\!\left(\dfrac{7}{4 - 4i}\right) = \tfrac12 \operatorname{Im}\!\left(\dfrac{7(4 + 4i)}{32}\right) = \tfrac12 \cdot \dfrac{28}{32} = \dfrac{7}{16}.$

Thus, B is the correct answer.

2020 AMC 12A Problem 22

Solution:

Problem 22 in Other Years