Question

在数列 $2001, 2002, 2003, \ldots$ 中，第三项之后每一项等于前面两项的和减去紧前一项。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：1460

解答：

递推式 $a_{k+1} = a_{k-2} + a_{k-1} - a_k$ 给出 $a_{k+1} - a_{k-1} = -(a_k - a_{k-2})$ 。数列开头为 $2001, 2002, 2003$ 、 $2000, 2005, 1998, \ldots$ 。

因此偶数位置项形成等差数列 $2002, 2000, 1998, \ldots$ ，公差为 $-2$ 。第 $2004$ 项是这个等差数列的第 $1002$ 项，即 $2002 + 1001(-2) = 0$ 。

所以正确答案是 C。

The recurrence $a_{k+1} = a_{k-2} + a_{k-1} - a_k$ gives $a_{k+1} - a_{k-1} = -(a_k - a_{k-2}).$ The sequence begins $2001, 2002, 2003,$ $2000, 2005, 1998, \ldots$

So the even-position terms form the arithmetic sequence $2002, 2000, 1998, \ldots$ with common difference $-2.$ The $2004$ th term is its $1002$ nd term, $2002 + 1001(-2) = 0.$

Thus, the correct answer is C.

2004 AMC 10B 第 19 题

答案：C

难度评级：1460

解答：

其他年份的第 19 题