Question

考虑形如 $x, 2000, y, \ldots$ 的正实数数列，其中从第二项开始，每一项都比它相邻两项的乘积少 $1$ 。对多少个不同的 $x$ 值，项 $2001$ 会出现在数列的某处？

Consider sequences of positive real numbers of the form $x, 2000, y, \ldots,$ in which every term after the first is $1$ less than the product of its two immediate neighbors. For how many different values of $x$ does the term $2001$ appear somewhere in the sequence?

$1$

$2$

$3$

$4$

多于 $4$

more than $4$

答案：D

知识点：递推分类讨论

难度评级：2390

解答：

若 $a, b, c$ 是连续三项，则 $b = ac - 1$ ，所以 $c = \dfrac{1 + b}{a}$ 。反复应用，前五项为之后 $a$ 和 $b$ 再次出现，因此数列以 $5$ 为周期。 $a,\ b,\ \dfrac{1 + b}{a},\ \dfrac{1 + a + b}{ab},\ \dfrac{1 + a}{b},$

此处 $b = 2000$ 是第二项。数值 $2001$ 可以放在五个不同位置中除第二项外的任意一个，每种选择都唯一确定 $2000$ 并产生一个有效的正实数数列。 $\begin{gathered} a=2001,\quad a=1,\\ a=\frac{2001}{4001999},\quad a=4001999. \end{gathered}$

所以共有 $4$ 个 $x$ 值。

因此，正确答案是 D。

If $a, b, c$ are consecutive terms then $b = ac - 1,$ so $c = \dfrac{1 + b}{a}.$ Applying this repeatedly, the first five terms are $a,\ b,\ \dfrac{1 + b}{a},\ \dfrac{1 + a + b}{ab},\ \dfrac{1 + a}{b},$ after which $a$ and $b$ recur, so the sequence is periodic with period $5.$

Here $b = 2000$ is the second term. Setting each of the other four displayed terms equal to $2001$ gives, respectively, $\begin{gathered} a=2001,\quad a=1,\\ a=\frac{2001}{4001999},\quad a=4001999. \end{gathered}$ These four values are positive and distinct. The second term itself is $2000,$ and periodicity shows there are no other positions to consider.

So there are $4$ values of $x.$

Thus, the correct answer is D.

2001 AMC 12 第 25 题

答案：D

难度评级：2390

解答：

其他年份的第 25 题