2001 AMC 12 第 25 题

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