AMC12 2013 A
AMC12 2013 A · Q7
AMC12 2013 A · Q7. It mainly tests Sequences & recursion (algebra).
The sequence $S_1, S_2, S_3, \dots, S_{10}$ has the property that every term beginning with the third is the sum of the previous two. That is, $S_n = S_{n-2} + S_{n-1}$ for $n \ge 3$. Suppose that $S_9 = 110$ and $S_7 = 42$. What is $S_4$?
数列 $S_1, S_2, S_3, \dots, S_{10}$ 具有从第三项开始，每一项是前两项之和的性质。即 $S_n = S_{n-2} + S_{n-1}$，$n \ge 3$。已知 $S_9 = 110$ 和 $S_7 = 42$。$S_4$ 是多少？
(A)
4
4
(B)
6
6
(C)
10
10
(D)
12
12
(E)
16
16
Answer
Correct choice: (C)
正确答案：（C）
Solution
Answer (C):
Note that $110=S_9=S_7+S_8=42+S_8$, so $S_8=110-42=68$. Thus $68=S_8=S_6+S_7=S_6+42$, so $S_6=68-42=26$. Similarly, $S_5=42-26=16$, and $S_4=26-16=10$.
答案（C）：
注意 $110=S_9=S_7+S_8=42+S_8$，所以 $S_8=110-42=68$。因此 $68=S_8=S_6+S_7=S_6+42$，所以 $S_6=68-42=26$。同理，$S_5=42-26=16$，并且 $S_4=26-16=10$。
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