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AMC 10 · 2024 · #23

Grade 5 arithmetic
Archetype Arithmetic Expression Decomposition
recursive-sequencesequences-geometricpattern-recognitionlucas-numbers pattern-recognitioneasier-related-problemidentify-subproblems ↑ Prerequisites: recursive-sequencesequences-arithmeticpattern-recognition
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Problem

The Fibonacci numbers are defined by F1=1,F2=1,F_1 = 1, F_2 = 1, and Fn=Fn−1+Fn−2F_n = F_{n-1} + F_{n-2} for n≥3.n \geq 3. What is F2F1+F4F2+F6F3+...+F20F10?{\frac{F_2}{F_1}} + {\frac{F_4}{F_2}} + {\frac{F_6}{F_3}} + ... + {\frac{F_{20}}{F_{10}}}?
(A) 318(B) 319(C) 320(D) 321(E) 322\textbf{(A) } 318 \qquad\textbf{(B) } 319 \qquad\textbf{(C) } 320 \qquad\textbf{(D) } 321 \qquad\textbf{(E) } 322

Pick an answer.

(A)
318
(B)
319
(C)
320
(D)
321
(E)
322
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Toolkit + CCSS Solution

Understand

Restated: The Fibonacci sequence is $F_1 = 1, F_2 = 1, F_n = F_{n-1} + F_{n-2}$. Compute the sum $\dfrac{F_2}{F_1} + \dfrac{F_4}{F_2} + \dfrac{F_6}{F_3} + \dots + \dfrac{F_{20}}{F_{10}}$ — that is, $\sum_{n=1}^{10} \dfrac{F_{2n}}{F_n}$.

Givens: $F_1 = F_2 = 1$; Recursion: $F_n = F_{n-1} + F_{n-2}$ for $n \ge 3$; Sum has $10$ terms: $n = 1, 2, \dots, 10$; Each term: $a_n = F_{2n}/F_n$; Answer choices: (A) $318$, (B) $319$, (C) $320$, (D) $321$, (E) $322$

Unknowns: The sum $S = a_1 + a_2 + \dots + a_{10}$

Understand

Restated: The Fibonacci sequence is $F_1 = 1, F_2 = 1, F_n = F_{n-1} + F_{n-2}$. Compute the sum $\dfrac{F_2}{F_1} + \dfrac{F_4}{F_2} + \dfrac{F_6}{F_3} + \dots + \dfrac{F_{20}}{F_{10}}$ — that is, $\sum_{n=1}^{10} \dfrac{F_{2n}}{F_n}$.

Givens: $F_1 = F_2 = 1$; Recursion: $F_n = F_{n-1} + F_{n-2}$ for $n \ge 3$; Sum has $10$ terms: $n = 1, 2, \dots, 10$; Each term: $a_n = F_{2n}/F_n$; Answer choices: (A) $318$, (B) $319$, (C) $320$, (D) $321$, (E) $322$

Plan

Primary tool: #2 Make a Systematic List

Secondary: #5 Look for a Pattern, #9 Solve an Easier Related Problem

The problem hands us a finite, well-defined sum of $10$ ratios. Tool #2 (Make a Systematic List) is the workhorse: list $F_1, \dots, F_{20}$ 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 ($1, 3, 4, 7, \dots$). 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.

Execute — Answer: B

#2 Make a Systematic List 4.OA.C.5 Step 1
  • List Fibonacci numbers $F_1$ through $F_{20}$ in order.
  • Each one is the sum of the previous two.
$$F_1, F_2, \dots, F_{20} = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765$$

💡 Grade 4 "generate a number pattern given a rule" — start with $1, 1$ and add.

#9 Solve an Easier Related Problem 5.NF.B.3 Step 2
  • Compute the first few ratios $a_n = F_{2n}/F_n$ explicitly.
  • $a_1 = F_2/F_1 = 1/1 = 1$.
  • $a_2 = F_4/F_2 = 3/1 = 3$.
  • $a_3 = F_6/F_3 = 8/2 = 4$.
  • $a_4 = F_8/F_4 = 21/3 = 7$.
  • $a_5 = F_{10}/F_5 = 55/5 = 11$.
  • So far: $1, 3, 4, 7, 11$.
$$a_1 = 1, \; a_2 = 3, \; a_3 = 4, \; a_4 = 7, \; a_5 = 11$$

💡 Grade 5 fraction-as-division: each $F_{2n}/F_n$ is a clean small integer because $F_n$ divides $F_{2n}$ exactly.

#5 Look for a Pattern 4.OA.C.5 Step 3
  • Spot the pattern.
  • $4 = 1 + 3, \; 7 = 3 + 4, \; 11 = 4 + 7$.
  • The sequence $a_n$ satisfies the same recurrence as Fibonacci, but starting from $1, 3$.
  • (This is the Lucas-number sequence, but no name is needed — the additive rule is enough.) Verify with one more term: $a_6$ should equal $7 + 11 = 18$, and indeed $F_{12}/F_6 = 144/8 = 18$.
  • Pattern confirmed.
$$a_n = a_{n-1} + a_{n-2} \text{ for } n \ge 3, \quad a_1 = 1, a_2 = 3$$

💡 Grade 4 number-pattern rule: compute four or five terms, conjecture, verify with one more — "$4 = 1 + 3$, $7 = 3 + 4$" is the standard tell.

#2 Make a Systematic List 4.OA.C.5 Step 4
  • Extend by addition to fill out all ten ratios.
  • $a_6 = 7 + 11 = 18$.
  • $a_7 = 11 + 18 = 29$.
  • $a_8 = 18 + 29 = 47$.
  • $a_9 = 29 + 47 = 76$.
  • $a_{10} = 47 + 76 = 123$.
  • (Spot check: $F_{20}/F_{10} = 6765/55 = 123$.
  • Match.)
$$a_6, a_7, a_8, a_9, a_{10} = 18, 29, 47, 76, 123$$

💡 Grade 4 pattern-extension by addition replaces five tedious Fibonacci divisions.

#2 Make a Systematic List 4.NBT.B.4 Step 5
  • Sum the ten ratios carefully.
  • Group from the small end: $1 + 3 = 4$; $4 + 4 = 8$; $8 + 7 = 15$; $15 + 11 = 26$; $26 + 18 = 44$; $44 + 29 = 73$; $73 + 47 = 120$; $120 + 76 = 196$; $196 + 123 = 319$.
  • So $S = 319$.
$$S = 1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = 319 \;\Rightarrow\; \textbf{(B)}$$

💡 Grade 4 multi-digit addition done in small chunks — partial sums grow steadily, no big number tricks needed.

[1] #2 4.OA.C.5 List Fibonacci numbers $F_1$ through $F_{20}$ in order. Each one is the sum of t
[2] #9 5.NF.B.3 Compute the first few ratios $a_n = F_{2n}/F_n$ explicitly. $a_1 = F_2/F_1 = 1/1
[3] #5 4.OA.C.5 Spot the pattern. $4 = 1 + 3, \; 7 = 3 + 4, \; 11 = 4 + 7$. The sequence $a_n$ s
[4] #2 4.OA.C.5 Extend by addition to fill out all ten ratios. $a_6 = 7 + 11 = 18$. $a_7 = 11 +
[5] #2 4.NBT.B.4 Sum the ten ratios carefully. Group from the small end: $1 + 3 = 4$; $4 + 4 = 8$

Review

Reasonableness: Two sanity checks. (1) Magnitude — the largest term is $a_{10} = 123$ and the second-largest is $a_9 = 76$, so the sum is at least $123 + 76 = 199$ and at most $10 \cdot 123 = 1230$; landing near $319$ (just above $123 \cdot 2 = 246$, well below $123 \cdot 4 = 492$) is sensible. (2) Recurrence reflection — sums of consecutive Lucas-style terms have a tidy closed form: $a_1 + a_2 + \dots + a_{10} = a_{12} - 3$. Computing $a_{11} = 76 + 123 = 199$ and $a_{12} = 123 + 199 = 322$ 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.

Alternative: Tool #14 (Use the recurrence as a telescoping identity). The identity $F_{2n} = F_n L_n$ (where $L_n$ are Lucas numbers) is well known, and $\sum_{n=1}^{N} L_n = L_{N+2} - 3$. Computing $L_{12}$ by the same $1, 3, 4, 7, \dots, 199, 322$ list, $S = 322 - 3 = 319$ in one line — no full per-term computation needed. This is the "closed form for the partial sum" approach; useful if you remember the identity, but our list-and-add path needs no remembered identity at all.

CCSS standards used (min grade 5)

  • 4.OA.C.5 Generate a number or shape pattern that follows a given rule. Identify apparent features of the pattern that were not explicit in the rule itself (Generating $F_1, \dots, F_{20}$ from the Fibonacci rule, then spotting that $a_n = F_{2n}/F_n$ obeys the same additive rule starting from $1, 3$, and extending by addition.)
  • 5.NF.B.3 Interpret a fraction as division of the numerator by the denominator; solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers (Reading $F_{2n}/F_n$ as plain division to compute $a_1 = 1, a_2 = 3, a_3 = 4, a_4 = 7, a_5 = 11$ — each a clean integer.)
  • 4.NBT.B.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm (Adding the ten ratios $1 + 3 + 4 + \dots + 123 = 319$ via small running partial sums.)

⭐ 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.

⭐ 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.

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