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) + … + F_20F_10?

2024 AMC 12B Problem 18

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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

Each term simplifies:

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

is an integer; compute the first few, such as

\dfrac{F_2}{F_1} = 1,

\dfrac{F_4}{F_2} = 3,

\dfrac{F_6}{F_3} = 4

These values are the Lucas numbers

L_k = 1, 3, 4, 7, 11, \ldots;

sum

L_1

through

L_{10}

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.