Question

递归定义数列： $F_{0}=0,~F_{1}=1$ ，且对所有 $n\geq 2$ ， $F_{n}=$ $F_{n-1}+F_{n-2}$ 除以 $3$ 的余数。因此数列开头为 $0,1,1,2,0,2,\ldots$ 。求 $F_{2017}+F_{2018}+F_{2019}+F_{2020}+$ $F_{2021}+F_{2022}+F_{2023}+F_{2024}?$

Define a sequence recursively by $F_{0}=0,~F_{1}=1,$ and $F_{n}=$ the remainder when $F_{n-1}+F_{n-2}$ is divided by $3,$ for all $n\geq 2.$ Thus the sequence starts $0,1,1,2,0,2,\ldots.$ What is $F_{2017}+F_{2018}+F_{2019}+F_{2020}+$ $F_{2021}+F_{2022}+F_{2023}+F_{2024}?$

$6$

$7$

$8$

$9$

$10$

答案：D

知识点：斐波那契数列模运算找规律

难度评级：1140

解答：

先列出前几项，看看能否找到数列中的规律。

$0, 1, 1, 2, 0, 2, 2, 1, 0, 1, \cdots$

由此可见，规律每 $8$ 项重复一次。 $(F_8,F_9)=(0,1)=(F_0,F_1)$ 。

所求是连续 $8$ 项的和，这个和固定。因此所以正确答案是 D。 $0 + 1 + 1 + 2 + 0 + 2$ $+ 2 + 1 = 9.$

Let us list out the first few values to see if we can find a pattern in this sequence.

$0, 1, 1, 2, 0, 2, 2, 1, 0, 1, \cdots$

The pair $(F_8,F_9)=(0,1)=(F_0,F_1).$ Because each later term is determined by the preceding two terms, the sequence therefore repeats every $8$ terms.

The desired answer is the sum of $8$ consecutive numbers, which is fixed. This sum is $0 + 1 + 1 + 2 + 0 + 2$ $+ 2 + 1 = 9.$ Thus, D is the correct answer.

2017 AMC 10A 第 13 题

答案：D

难度评级：1140

解答：

其他年份的第 13 题