The Fibonacci numbers are defined by F_1 = 1, F_2 = 1, and F_n = F_n-1 + F_n-2 for n ≥ 3. What is (F_2)/(F_1) + (F_4)/(F_2) + (F_6)/(F_3) + \cdots + \fracF_20F_10?

2024 AMC 12B Problem 18 — Solution

2024 AMC 12B Problem 18

Attempt Problem 18 of the 2024 AMC 12B 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 2024 AMC 12B solutions, or check the answer key.

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18.

The Fibonacci numbers are defined by $F_1 = 1,$ $F_2 = 1,$ and $F_n = F_{n-1} + F_{n-2}$ for $n \ge 3.$ What is

$\frac{F_2}{F_1} + \frac{F_4}{F_2} + \frac{F_6}{F_3} + \cdots + \frac{F_{20}}{F_{10}}?$

$318$

$319$

$320$

$321$

$322$

Answer: B

Concepts:Fibonacci summation

Difficulty rating: 1930

Solution:

Since $F_{2k} = F_k L_k$ where $L_k$ is the $k$ th Lucas number, each term $\dfrac{F_{2k}}{F_k} = L_k.$ The sum is $\begin{gathered} L_1 + L_2 + \cdots + L_{10} \\ = 1 + 3 + 4 + 7 + 11 + 18 \\ {}+ 29 + 47 + 76 + 123 \\ = 319. \end{gathered}$ (Equivalently, $L_1 + \cdots + L_{10}$ $= L_{12} - 3$ $= 322 - 3 = 319.$ )

Thus, the correct answer is B.

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