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

(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}

Answer: B

Concepts:complex number summation

Difficulty rating: 2110

Small hint:

a_n b_n

= \tfrac12 \operatorname{Im}\big((a_n + b_n i)^2\big)

= \tfrac12 \operatorname{Im}\big((2 + i)^{2n}\big)

Big hint:

Since

(2 + i)^2 = 3 + 4i,

the sum becomes a geometric series with ratio

\dfrac{3 + 4i}{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.

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