AMC 10 · 2024 · #18

Grade 5 arithmetic
recursive-sequencesequences-geometricpattern-recognitionlucas-numbers pattern-recognitioneasier-related-problemidentify-subproblems ↑ Prerequisites: recursive-sequencesequences-arithmeticpattern-recognition
📏 Medium solution 💡 3 insights
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Problem
The Fibonacci sequence is F₁ = 1, F₂ = 1, F_n = F_n-1 + F_n-2. Compute the sum F₂/F₁ + F₄/F₂ + F₆/F₃ + … + F₂₀/F₁₀ — that is, Σ_n=1¹⁰ F₂n/F_n.

Pick an answer.

(A)
318
(B)
319
(C)
320
(D)
321
(E)
322
How to solve
Strategy Make a Systematic List

The problem hands us a finite, well-defined sum of 10 ratios. Tool #2 (Make a Systematic List) is the workhorse: list F₁, …, F₂₀ in order, then list the 10 ratios. Tool #9 (Solve an Easier Related Problem) gets us moving even before that — compute the first three or four ratios by hand and notice they are integers (1, 3, 4, 7, …). Tool #5 (Look for a Pattern) then turns those into a Fibonacci-style recurrence (a_n = a_n-1 + a_n-2 starting 1, 3), so the rest of the list extends by addition rather than by dividing big Fibonacci numbers. With 10 integers in hand, the answer is one careful sum.

1STEP 1

List the terms you need

Write out the first twenty terms.

F₁, F₂, …, F₂₀ = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765
2STEP 2

Compute the first few quotients

They come out 1, 3, 4, 7, 11.

a₁ = 1, a₂ = 3, a₃ = 4, a₄ = 7, a₅ = 11
3STEP 3

Spot the rule

Each quotient is the sum of the previous two.

a_n = a_n-1 + a_n-2 for n ≥ 3, a₁ = 1, a₂ = 3
4STEP 4

Continue the list

Next come 18, 29, 47, 76, 123.

a₆, a₇, a₈, a₉, a₁₀ = 18, 29, 47, 76, 123
5STEP 5

Add the ten terms

Their total is 319.

S = 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = 319 → (B)
Answer
319
Two sanity checks. (1) Magnitude — the largest term is a₁₀ = 123 and the second-largest is a₉ = 76, so the sum is at least 123 + 76 = 199 and at most 10 · 123 = 1230; landing near 319 (just above 123 · 2 = 246, well below 123 · 4 = 492) is sensible. (2) Recurrence reflection — sums of consecutive Lucas-style terms have a tidy closed form: a₁ + a₂ + … + a₁₀ = a₁₂ - 3. Computing a₁₁ = 76 + 123 = 199 and a₁₂ = 123 + 199 = 322 gives 322 - 3 = 319, which matches and uses the same recurrence as a free verification. The nearby distractors 318, 320, 321, 322 are exactly the off-by-one/off-by-two arithmetic traps this re-derivation rules out.
💡Key takeaway

Hard-looking Fibonacci ratios become a simple pattern game once you compute the first few: 1, 3, 4, 7, 11 extends by addition to 123, and ten numbers sum to 319 — just Grade 5 division and Grade 4 number-pattern rules.