2007 AIME I Problems/Problem 14: Difference between revisions

Revision as of 19:11, 4 July 2013

Problem

A sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ , $a_{n+1}a_{n-1}=a_{n}^{2}+2007$ .

Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}$

Solution

We are given that

$a_{n+1}a_{n-1}= a_{n}^{2}+2007$ , $a_{n-1}^{2}+2007 = a_{n}a_{n-2}$ .

Add these two equations to get

$a_{n-1}(a_{n-1}+a_{n+1}) = a_{n}(a_{n}+a_{n-2})$

$\frac{a_{n+1}+a_{n-1}}{a_{n}}= \frac{a_{n}+a_{n-2}}{a_{n-1}}$ .

This is an invariant. Defining $b_{i}= \frac{a_{i}}{a_{i-1}}$ for each $i \ge 2$ , the above equation means

$b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}$ .

We can thus calculate that $b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225$ . Now notice that $b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}> \frac{a_{2006}}{a_{2005}}= b_{2006}$ . This means that

$b_{2007}+\frac{1}{b_{2007}}< b_{2007}+\frac{1}{b_{2006}}= 225$ . It is only a tiny bit less because all the $b_i$ are greater than $1$ , so we conclude that the floor of $\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}$ is $224$ .

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	[[Category:Intermediate Algebra Problems]]		[[Category:Intermediate Algebra Problems]]
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2007 AIME I Problems/Problem 14: Difference between revisions

Revision as of 19:11, 4 July 2013

Problem

Solution

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