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

小提示：

条件 $b = ac - 1$ 可整理为 $c = \dfrac{1 + b}{a}$ ，这让你可以向后逐项推进

The condition $b = ac - 1$ rearranges to $c = \dfrac{1 + b}{a},$ letting you march forward

大提示：

反复这样做可知数列以 $5$ 为周期，所以 $2001$ 只能是五个位置之一

Doing this repeatedly shows the sequence is periodic with period $5,$ so $2001$ can only be one of five terms

解答：

若 $a, b, c$ 是连续三项，则 $b = ac - 1$ ，所以 $c = \dfrac{1 + b}{a}$ 。反复应用这个关系，前五项依次为 $a,\ b,\ \dfrac{1 + b}{a},\ \dfrac{1 + a + b}{ab},\ \dfrac{1 + a}{b}\text{，}$ 此后 $a$ 和 $b$ 又重新出现，因此数列以 $5$ 为周期。

此处 $b = 2000$ 是第二项。令上面另外四项分别等于 $2001$ ，依次解得 $\begin{gathered} a=2001,\quad a=1,\\ a=\frac{2001}{4001999},\quad a=4001999 \end{gathered}\text{。}$ 这四个值都是正数且互不相同。而第二项本身等于 $2000$ ，再由周期性可知没有其他位置需要考虑。

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

小提示：

大提示：

解答：

其他年份的第 25 题