AMC 10 · 2024 · #23
Grade 5 arithmeticProblem
The Fibonacci numbers are defined by F1=1,F2=1, and Fn=Fn−1+Fn−2 for n≥3. What is F1F2+F2F4+F3F6+...+F10F20?
Pick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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.
💡 Grade 4 "generate a number pattern given a rule" — start with $1, 1$ and add.
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: $1, 3, 4, 7, 11$.
💡 Grade 5 fraction-as-division: each $F_{2n}/F_n$ is a clean small integer because $F_n$ divides $F_{2n}$ exactly.
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.
💡 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.
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.)
💡 Grade 4 pattern-extension by addition replaces five tedious Fibonacci divisions.
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 $S = 319$.
💡 Grade 4 multi-digit addition done in small chunks — partial sums grow steadily, no big number tricks needed.
4.OA.C.5 List Fibonacci numbers $F_1$ through $F_{20}$ in order. Each one is the sum of t 5.NF.B.3 Compute the first few ratios $a_n = F_{2n}/F_n$ explicitly. $a_1 = F_2/F_1 = 1/1 4.OA.C.5 Spot the pattern. $4 = 1 + 3, \; 7 = 3 + 4, \; 11 = 4 + 7$. The sequence $a_n$ s 4.OA.C.5 Extend by addition to fill out all ten ratios. $a_6 = 7 + 11 = 18$. $a_7 = 11 + 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.5Generate 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.3Interpret 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.4Fluently 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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