AMC10 2004 B

AMC10 2004 B · Q19

AMC10 2004 B · Q19. It mainly tests Sequences & recursion (algebra), Patterns & sequences (misc).

In the sequence 2001, 2002, 2003, \dots, 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 2004th term in this sequence?

在数列2001, 2002, 2003, \dots中，从第四项起，每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？

(A) -2004 -2004

(B) -2 -2

(D) 4003 4003

(E) 6007 6007

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) Let $a_k$ be the $k$th term of the sequence. For $k \ge 3$, $a_{k+1} = a_{k-2} + a_{k-1} - a_k,$ so $a_{k+1} - a_{k-1} = -(a_k - a_{k-2}).$ Because the sequence begins $2001, 2002, 2003, 2000, 2005, 1998, \ldots,$ it follows that the odd-numbered terms and the even-numbered terms each form arithmetic progressions with common differences of $2$ and $-2$, respectively. The $2004$th term of the original sequence is the $1002$nd term of the sequence $2002, 2000, 1998, \ldots,$ and that term is $2002 + 1001(-2) = 0$.

（C）设 $a_k$ 为该数列的第 $k$ 项。对 $k \ge 3$， $a_{k+1} = a_{k-2} + a_{k-1} - a_k,$ 因此 $a_{k+1} - a_{k-1} = -(a_k - a_{k-2}).$ 由于数列开始为 $2001, 2002, 2003, 2000, 2005, 1998, \ldots,$ 可知奇数项与偶数项分别构成等差数列，其公差分别为 $2$ 和 $-2$。原数列的第 $2004$ 项是数列 $2002, 2000, 1998, \ldots,$ 的第 $1002$ 项，该项为 $2002 + 1001(-2) = 0$。

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