AMC 10 · 2020 · #22

Grade 11 algebra

Archetype Geometric Series Share └ Sub-archetype Cyclic Participant Share

complex-numbersgeometric-series-infinitesequences-geometricexponents convert-to-algebrapattern-recognitionidentify-subproblems ↑ Prerequisites: complex-numbersgeometric-series-infinite

📏 Long solution 💡 3 insights

Problem

Expanding a fixed complex number to the power n always leaves a complex number; its real and imaginary parts define two sequences at once. Multiply the two parts together, divide by seven to the n, and add over every n from zero upward. Find the value of that infinite sum.

Pick an answer.

(A)

$\frac 38$

(B)

$\frac7{16}$

(C)

$\frac12$

(D)

$\frac9{16}$

(E)

$\frac47$

How to solve

Strategy Organize Information in More Ways

Tool #15 (Organize Information in More Ways) is the whole problem. Chasing a_n and b_n separately is hopeless — they satisfy a coupled recurrence and neither has a clean closed form. But the product a_n b_n is exactly half of the imaginary part of (a_n + b_n i)², because (a+bi)² = (a² - b²) + 2ab i. That single re-packaging turns a product of two unknown sequences into one imaginary part of one complex power. Tool #4 (Introduce a Variable): name z = (2+i)² = 3+4i so the powers collapse. Tool #5 (Look for a Pattern): once the summand is (z/7)ⁿ, the series is plainly geometric. Tool #6 (Guess and Check): add the first several terms numerically to confirm the closed form lands on the right answer choice.

1STEP 1

Repackage the product

The product is the imaginary part of a square.

(a_n + b_n i)² = (a_n² - b_n²) + 2 a_n b_n i ⟹ a_n b_n = 1/2Im((2+i)²ⁿ)

2STEP 2

Collapse the even powers

Take the square as a new base.

z = (2+i)² = 3+4i, a_n b_n = 1/2Im(zⁿ)

3STEP 3

Pull the sum inside

Move the sum inside the imaginary part.

Σ_n=0^∞(a_n b_n)/7ⁿ = 1/2Σ_n=0^∞Im(rⁿ) = 1/2Im(Σ_n=0^∞ rⁿ), r = (3+4i)/7

4STEP 4

Sum the geometric series

The modulus is under one, so it converges.

|r| = 5/7 ＜ 1, Σ_n=0^∞ rⁿ = 1/(1-r) = 7/(4-4i)

5STEP 5

Divide using the conjugate

The conjugate clears the denominator.

7/(4-4i) = (7(4+4i))/(4-4i)(4+4i) = (28+28i)/32 = 7/8 + 7/8i

6STEP 6

Read the imaginary part and halve

Halving gives seven sixteenths.

Σ_n=0^∞(a_n b_n)/7ⁿ = 1/2·7/8 = 7/16 → (B)

Answer

7/16

Numerical check with partial sums. From (2+i)⁰ = 1, (2+i)¹ = 2+i, (2+i)² = 3+4i, (2+i)³ = 2+11i, (2+i)⁴ = -7+24i, (2+i)⁵ = -38+41i, (2+i)⁶ = -117+44i, the terms (a_n b_n)/7ⁿ are 0, 2/7 ≈ 0.286, 12/49 ≈ 0.245, 22/343 ≈ 0.064, -168/2401 ≈ -0.070, ≈ -0.093, ≈ -0.044. The running totals go 0, 0.286, 0.531, 0.595, 0.525, 0.432, 0.388, and keep oscillating inward toward 0.4375 = 7/16 — matching (B) and ruling out (A) 0.375, (C) 0.5, (D) 0.5625, (E) ≈ 0.571. The oscillation is expected: 3+4i has argument well away from 0, so Im(zⁿ) changes sign, and the envelope shrinks like (5/7)ⁿ exactly as the convergence test predicted. Structural check: the answer is positive and under 1/2, consistent with the first positive terms being partly cancelled by later negative ones.

💡Key takeaway

When two unknown sequences appear only as a product, square the thing that made them: (a+bi)² = (a²-b²) + 2abi parks the product ab in the imaginary part, and here that turns the whole sum into one geometric series with ratio (3+4i)/7.

Repackage the product as one imaginary part
Collapse the even powers
Pull the sum inside Im
Sum the geometric series
Divide using the conjugate
Read the imaginary part and halve