AMC12 2024 B

AMC12 2024 B · Q18

AMC12 2024 B · Q18. It mainly tests Sequences & recursion (algebra), Basic counting (rules of product/sum).

The Fibonacci numbers are defined by $F_1 = 1, F_2 = 1,$ and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 3.$ What is \[{\frac{F_2}{F_1}} + {\frac{F_4}{F_2}} + {\frac{F_6}{F_3}} + ... + {\frac{F_{20}}{F_{10}}}?\]

斐波那契数列定义为 $F_1 = 1, F_2 = 1,$ 且 $F_n = F_{n-1} + F_{n-2}$ 对于 $n \geq 3$。求 \[{\frac{F_2}{F_1}} + {\frac{F_4}{F_2}} + {\frac{F_6}{F_3}} + ... + {\frac{F_{20}}{F_{10}}}?\]

(A) 318 318

(B) 319 319

(D) 321 321

(E) 322 322

Answer

Correct choice: (B)

正确答案：（B）

Solution

The first $20$ terms are $F_n = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765$ So the answer is $1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = \boxed{(B) 319}$. - Do not do this unless you have no other option or no time to find a smarter way

前 $20$ 项为 $F_n = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765$ 因此答案是 $1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = \boxed{(B) 319}$。 - 除非没有其他选择或没有时间找到更聪明的方法，否则不要这样做

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