2024 AMC 12B Problema 18

Question

Los números de Fibonacci se definen por

F_1 = 1,

F_2 = 1,

y

F_n = F_{n-1} + F_{n-2}

para

n \ge 3.

¿Cuánto vale

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

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

Respuesta: B

Conceptos:Fibonacci sumatoria

Nivel de dificultad: 1930

Pista pequeña:

Cada término se simplifica:

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

es un entero; calcula los primeros, como

\dfrac{F_2}{F_1} = 1,

\dfrac{F_4}{F_2} = 3,

\dfrac{F_6}{F_3} = 4

.

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

Pista grande:

Estos valores son los números de Lucas

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

suma

L_1

hasta

L_{10}

.

These values are the Lucas numbers

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

sum

L_1

through

L_{10}

Solución:

Como

F_{2k} = F_k L_k

donde

L_k

es el

k

-ésimo número de Lucas, cada término

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

La suma es

\begin{gathered} L_1 + L_2 + \cdots + L_{10} \\ = 1 + 3 + 4 + 7 + 11 + 18 \\ {}+ 29 + 47 + 76 + 123 \\ = 319. \end{gathered}

(Equivalentemente,

L_1 + \cdots + L_{10}

= L_{12} - 3

= 322 - 3 = 319.

)

Por lo tanto, la respuesta correcta es B.

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.