2024 AMC 10B Problem 23

Attempt Problem 23 of the 2024 AMC 10B 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 10B solutions, or check the answer key.

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23.

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: 2270

Small hint:

Each term is

\dfrac{F_{2k}}{F_k};

use the identity

F_{2k} = F_k L_k,

where

L_k

is the Lucas number

Big hint:

The sum becomes

L_1 + L_2 + \cdots + L_{10};

sums of Lucas numbers satisfy

\sum_{k=1}^{n} L_k = L_{n+2} - 3

Solution:

Use

F_{2k} = F_k L_k,

so each term

\dfrac{F_{2k}}{F_k} = L_k,

the

k

th Lucas number. That collapses the sum to

\sum_{k=1}^{10} L_k.

With

L_1 = 1,

L_2 = 3,

L_3 = 4,

\ldots,

L_{10} = 123,

the identity

\sum_{k=1}^{n} L_k = L_{n+2} - 3

gives

L_{12} - 3 = 322 - 3 = 319.

Thus, B is the correct answer.