2020 AMC 12A Problem 22

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.