Question

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，已知 $S_9 = 110$ 且 $S_7 = 42$ 。 $S_4$ 是多少？ $S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3.$

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

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.

2013 AMC 12A 第 7 题

答案：C

难度评级：1270

解答：

其他年份的第 7 题