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Difference between revisions of "2007 AIME I Problems/Problem 14"

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Add these two equations to get
 
Add these two equations to get
  
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:<math>\displaystyle a_{n-1}(a_{n-1}+a_{n+1}) = a_{n}(a_{n}+a_{n-2})</math>
+
:<math>a_{n-1}(a_{n-1}+a_{n+1}) = a_{n}(a_{n}+a_{n-2})</math>
  
 
:<math>\frac{a_{n+1}+a_{n-1}}{a_{n}}= \frac{a_{n}+a_{n-2}}{a_{n-1}}</math>.
 
:<math>\frac{a_{n+1}+a_{n-1}}{a_{n}}= \frac{a_{n}+a_{n-2}}{a_{n-1}}</math>.
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We can thus calculate that <math>b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225</math>. Now notice that <math>b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}> \frac{a_{2006}}{a_{2005}}= b_{2006}</math>. This means that
 
We can thus calculate that <math>b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225</math>. Now notice that <math>b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}> \frac{a_{2006}}{a_{2005}}= b_{2006}</math>. This means that
  
−
<math>b_{2007}+\frac{1}{b_{2007}}< b_{2007}+\frac{1}{b_{2006}}= 225</math>. But since it's only a tiny bit less, we conclude that the floor of <math>\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}</math> is <math>224</math>.
+
<math>b_{2007}+\frac{1}{b_{2007}}< b_{2007}+\frac{1}{b_{2006}}= 225</math>. It is only a tiny bit less because all the <math>b_i</math> are greater than <math>1</math>, so we conclude that the floor of <math>\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}</math> is <math>224</math>.
  
 
== See also ==
 
== See also ==

Revision as of 17:09, 19 January 2008

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 $224$.

See also

2007 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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