Question

Consider sequences of positive real numbers of the form \( x, 2000, y, \dots \), 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?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

In recurrence relations, systematically check each case to determine the number of solutions that satisfy the condition.

Answer 1

The Correct Option is D

We have the sequence \( x, 2000, y, \dots \), and the recurrence relation for each term is given by: \[ a_n = a_{n-1} \cdot a_{n+1} - 1 \] For \( 2001 \) to appear in the sequence, solving the recurrence relations will give 4 possible values for \( x \).