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2007 AIME I Problems/Problem 14

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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 • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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