AMC12 2004 B

AMC12 2004 B · Q12

AMC12 2004 B · Q12. It mainly tests Functions basics, Sequences & recursion (algebra).

In the sequence $2001$, $2002$, $2003$, $\ldots$ , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000$. What is the $2004^\textrm{th}$ term in this sequence?

在数列$2001$, $2002$, $2003$, $\ldots$中，从第四项开始，每一项由其前两项之和减去前一项得到。例如，第四项是$2001 + 2002 - 2003 = 2000$。该数列的第$2004^\textrm{th}$项是多少？

(A) -2004 -2004

(B) -2 -2

(D) 4003 4003

(E) 6007 6007

Answer

Correct choice: (C)

正确答案：（C）

Solution

We already know that $a_1=2001$, $a_2=2002$, $a_3=2003$, and $a_4=2000$. Let's compute the next few terms to get the idea how the sequence behaves. We get $a_5 = 2002+2003-2000 = 2005$, $a_6=2003+2000-2005=1998$, $a_7=2000+2005-1998=2007$, and so on. We can now discover the following pattern: $a_{2k+1} = 2001+2k$ and $a_{2k}=2004-2k$. This is easily proved by induction. It follows that $a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}$.

已知$a_1=2001$, $a_2=2002$, $a_3=2003$, $a_4=2000$。计算接下来的几项以观察规律： $a_5 = 2002+2003-2000 = 2005$, $a_6=2003+2000-2005=1998$, $a_7=2000+2005-1998=2007$，等等。由此可发现如下模式：$a_{2k+1} = 2001+2k$且$a_{2k}=2004-2k$。这可用归纳法轻易证明。于是 $a_{2004}=a_{2\cdot 1002} = 2004 - 2\cdot 1002 = \boxed{0}$。

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