2009 AMC 8 第 5 题

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5.

一个数列以 112233 开始。数列的第四个数是前三个数之和,即 1+2+3=61 + 2 + 3 = 6。同样地,第四个数之后的每个数都是它前面三个数之和。这个数列的第八个数是多少?

A sequence of numbers starts with 1,1, 2,2, and 3.3. The fourth number of the sequence is the sum of the previous three numbers in the sequence: 1+2+3=6.1 + 2 + 3 = 6. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence?

1111

2020

3737

6868

9999

答案:D
知识点:递推
难度评级:770
解答:

数列开始为 1,2,31,2,3

接下来的项为 1+2+3=61+2+3=62+3+6=112+3+6=113+6+11=203+6+11=206+11+20=376+11+20=3711+20+37=6811+20+37=68

所以正确答案是 D

The sequence begins 1,2,31,2,3.

The next terms are 1+2+3=61+2+3=6, 2+3+6=112+3+6=11, 3+6+11=203+6+11=20, 6+11+20=376+11+20=37, and 11+20+37=6811+20+37=68.

Thus, D is the correct answer.

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