Question

在数列 $2001$ ， $2002$ ， $2003$ ， $\ldots$ 中，从第四项起，每一项等于前两项之和减去前一项。例如第四项为 $2001 + 2002 - 2003 = 2000$ 。这个数列的第 $2004$ 项是多少？

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 $2004$ th term in this sequence?

$-2004$

$-2$

$0$

$4003$

$6007$

答案：C

知识点：递推等差数列找规律

难度评级：1500

小提示：

写出若干项来寻找规律。

Write out several terms to spot a pattern

大提示：

偶数项 $2002, 2000, 1998, \ldots$ 构成等差数列。

The even-indexed terms $2002, 2000, 1998, \ldots$ form an arithmetic progression

解答：

递推给出 $2001$ 、 $2002$ 、 $2003$ 、 $2000$ 、 $2005$ 、 $1998, \ldots$ 。偶数项为 $2002, 2000, 1998, \ldots$ ，每次减少 $2$ 。

更确切地说，用归纳法可以验证 $a_{2k} = 2004 - 2k$ 且 $a_{2k+1} = 2001 + 2k$ 。因此 $\begin{aligned} a_{2004} &= a_{2\cdot1002} \\ &= 2004 - 2(1002) = 0 \end{aligned}\text{。}$

因此正确答案是 C。

The rule gives $2001,$ $2002,$ $2003,$ $2000,$ $2005,$ $1998, \ldots$ The even-indexed terms are $2002, 2000, 1998, \ldots,$ decreasing by $2.$

More precisely, the recurrence verifies inductively that $a_{2k} = 2004 - 2k$ and $a_{2k+1} = 2001 + 2k.$ Therefore $\begin{aligned} a_{2004} &= a_{2\cdot1002} \\ &= 2004 - 2(1002) = 0. \end{aligned}$

Thus, the correct answer is C.

2004 AMC 12B 第 12 题

答案：C

难度评级：1500

小提示：

大提示：

解答：

其他年份的第 12 题