2013 AMC 12A 第 7 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

7.

数列

S_1

，

S_2

，

S_3

，

\ldots

，

S_{10}

满足从第三项开始，每一项都是前两项之和。也就是说，

S_n = S_{n-2} + S_{n-1} \text{ 对所有 } n \ge 3\text{。}

已知

S_9 = 110

且

S_7 = 42

。

S_4

是多少？

The sequence

S_1,

S_2,

S_3,

\ldots,

S_{10}

has the property that every term beginning with the third is the sum of the previous two. That is,

S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3.

Suppose that

S_9 = 110

and

S_7 = 42.

What is

S_4?

4

6

10

12

16

答案：C

知识点：递推逆推法

难度评级：1270

小提示：

将递推式改写为

S_{n-2} = S_n - S_{n-1}

，然后向前倒推

Rearrange the rule to

S_{n-2} = S_n - S_{n-1}

and step downward

大提示：

由

S_9 = S_7 + S_8

求出

S_8

，再倒推到

S_4

。

From

S_9 = S_7 + S_8

find

S_8,

then work down to

S_4

解答：

因为

S_9 = S_7 + S_8

，所以

S_8 = 110 - 42 = 68

。然后

S_6 = S_8 - S_7 = 68 - 42 = 26

，

S_5 = S_7 - S_6 = 42 - 26 = 16

，且

S_4 = S_6 - S_5 = 26 - 16 = 10

。

因此，正确答案是 C。

Since

S_9 = S_7 + S_8,

we get

S_8 = 110 - 42 = 68.

Then

S_6 = S_8 - S_7 = 68 - 42 = 26,

S_5 = S_7 - S_6 = 42 - 26 = 16,

and

S_4 = S_6 - S_5 = 26 - 16 = 10.

Thus, the correct answer is C.

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