2020 AMC 12A Problem 22

Attempt Problem 22 of the 2020 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 12A solutions, or check the answer key.

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22.

Let (an)(a_n) and (bn)(b_n) be the sequences of real numbers such that (2+i)n=an+bni(2 + i)^n = a_n + b_n i for all integers n0,n \ge 0, where i=1.i = \sqrt{-1}. What is n=0anbn7n?\sum_{n=0}^{\infty} \frac{a_n b_n}{7^n}?

38\dfrac{3}{8}

716\dfrac{7}{16}

12\dfrac{1}{2}

916\dfrac{9}{16}

47\dfrac{4}{7}

Answer: B
Concepts:complex numbersummation
Difficulty rating: 2110
Solution:

Since (an+bni)2=an2bn2+2anbni,(a_n + b_n i)^2 = a_n^2 - b_n^2 + 2 a_n b_n i, we have anbna_n b_n =12Im((2+i)2n)= \tfrac12 \operatorname{Im}\big((2 + i)^{2n}\big) =12Im((3+4i)n).= \tfrac12 \operatorname{Im}\big((3 + 4i)^n\big).

Therefore the sum is 12Imn=0(3+4i7)n\tfrac12 \operatorname{Im} \displaystyle\sum_{n=0}^{\infty} \left(\frac{3 + 4i}{7}\right)^n =12Im ⁣(113+4i7).= \tfrac12 \operatorname{Im}\!\left(\frac{1}{1 - \frac{3 + 4i}{7}}\right).

This equals 12Im ⁣(744i)\tfrac12 \operatorname{Im}\!\left(\dfrac{7}{4 - 4i}\right) =12Im ⁣(7(4+4i)32)= \tfrac12 \operatorname{Im}\!\left(\dfrac{7(4 + 4i)}{32}\right) =122832= \tfrac12 \cdot \dfrac{28}{32} =716.= \dfrac{7}{16}.

Thus, B is the correct answer.

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