Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

"Consider sequences of positive real numbers of the form x,2000,y,..., 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 (E) More than 4"

Can anyone suggest a systematic way to solve this problem? Thanks!

share|improve this question

4 Answers 4

up vote 5 down vote accepted

Perhaps a bit of a tedious solution but this is the most straight forward approach. By direct substitution it is not difficult to show that the sequence necessarily repeats itself after the $6$th term. We have $$a_1 = x$$ $$a_2 = 2000$$ $$a_3 = \frac{2001}{x}$$ $$a_4 = \frac{2001 + x}{2000x}$$ $$a_5 = \frac{1 + x}{2000}$$ $$a_6 = x$$ $$a_7 = 2000$$ and so on.... Clearly we can take $x=2001$ for a trivial appearance. For $a_3$ to be $2001$ we require $x=1$. For $a_4$ we require $x=0.0005$. For $a_5$ to be $2001$ we require $x = 4001999$. This gives a total of $4$ solutions for $x$.

share|improve this answer

You have a sequence defined by $a_0=x,a_1=2000$, and the recurrence $$a_{n+1}=\frac{a_n+1}{a_{n-1}}\;.$$ Calculate the first few terms:

$$\begin{align*} a_0&=x\\\\ a_1&=2000\\\\ a_2&=\frac{2001}x\\\\ a_3&=\frac{2001+x}{2000x}\\\\ a_4&=\frac{\frac{2001(x+1)}{2000x}}{\frac{2001}{x}}=\frac{x+1}{2000}\\\\ a_5&=\frac{\frac{x+2001}{2000}}{\frac{x+2001}{2000x}}=x\\\\ a_6&=2000 \end{align*}$$

Clearly the sequence is period with period $5$, so $2001$ appears iff

$$2001\in\left\{x,\frac{2001}x,\frac{x+2001}{2000x},\frac{x+1}{2000}\right\}\;.$$

Thus, there are four solutions: $2001,1,\dfrac{4001999}{2001}$, and $4001999$.

share|improve this answer
    
@Robert: Indeed, that becomes pretty apparent as soon as one looks closely at the specific calculations. –  Brian M. Scott Aug 15 '12 at 6:16

(This is basically EuYu's answer with the details of periodicity added; took a while to type up.)

Suppose that $a_0 , a_1 , \ldots$ is a generalised sequence of the type described, so that $a_i = a_{i-1} a_{i+1} - 1$ for all $i > 0$. Note that this condition is equivalent to demanding that $$a_{i+1} = \frac{ a_i + 1 }{a_{i-1}}.$$

Using this we find the following recurrences: $$ a_2 = \frac{ a_1 + 1}{a_0}; \\ a_3 = \frac{ a_2 + 1}{a_1} = \frac{ \frac{ a_1 + 1}{a_0} }{a_1} = \frac{ a_0 + a_1 + 1 }{ a_0a_1 }; \\ a_4 = \frac{ a_3 + 1 }{a_2} = \frac{\frac{ a_0 + a_1 + 1 }{ a_0a_1 } + 1}{\frac{ a_1 + 1}{a_0}} = \frac{ ( a_0 + 1 )( a_1 + 1) }{ a_1 ( a_1 + 1 ) } = \frac{a_0 + 1}{a_1};\\ a_5 = \frac{ a_4 + 1 }{ a_3 } = \frac{ \frac{a_0 + 1}{a_1} + 1}{\frac{ a_0 + a_1 + 1 }{ a_0a_1 }} = \frac{ \left( \frac{a_0 + a_1 + 1}{a_1} \right) }{ \left( \frac{a_0+a_1+1}{a_0a_1} \right) } = a_0 \\ a_6 = \frac{ a_5 + 1 }{a_4} = \frac{ a_0 + 1}{ \left( \frac{ a_0 + 1 }{a_1} \right) } = a_1. $$

Thus every such sequence is periodic with period 5, so if 2001 appears, it must appear as either $a_0, a_1, a_2, a_3, a_4$.

  1. Clearly if $a_0 = 2001$, we're done.
  2. As we stipulate that $a_1 = 2000$, it is impossible for $a_1 = 2001$.
  3. If $a_2 = 2001$, then it must be that $2001 = \frac{ 2000 + 1 }{a_0}$ and so $a_0 = 1$.
  4. If $a_3 = 2001$, then it must be that $2001 = \frac{a_0 + 2000 + 1}{a_0 \cdot 2000}$, and it follows that $a_0 = \frac{2001}{2000 \cdot 2001 - 1}$.
  5. If $a_4 = 2001$, then it must be that $2001 = \frac{ a_0 + 1 }{2000}$, and so $a_0 = 2001 \cdot 2000 - 1$.

There are thus exactly four values of $a_0$ such that 2001 appears in the sequence.

share|improve this answer

We are given that $xy=2001$. Clearly if one of them is $1$ and the other is $2001$, then $2001$ appears. Try the other factorizations to see if $2001$ appears.

share|improve this answer
    
The problem specifies real numbers - are you assuming they're all integers? –  Gerry Myerson Aug 15 '12 at 4:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.