2007 AIME I Problems/Problem 14

Revision as of 19:41, 25 January 2014 by Mathgeek2006 (talk | contribs)


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 1

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 $224$.

Solution 2

The equation $a_{n+1}a_{n-1}-a_n^2=2007$ looks like the determinant \[\left|\begin{array}{cc}a_{n+1}&a_n\\a_n&a_{n-1}\end{array}\right|=2007.\] Therefore, the determinant of this matrix is invariant. Guessing that this sequence might be a linear recursion because of the matrix form given below, we define the sequence $b_n$ defined by $b_1=b_2=3$ and $b_{n+1}=\alpha b_n+\beta b_{n-1}$ for $n\ge 2$. We wish to find $\alpha$ and $\beta$ such that $a_n=b_n$ for all $n\ge 1$. To do this, we use the following matrix form of a linear recurrence relation


When we take determinants, this equation becomes


We want \[\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=2007\] for all $n$. Therefore, we replace the two matrices by $2007$ to find that

\[2007=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\cdot 2007\] \[1=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)=-\beta.\] Therefore, $\beta=-1$. Computing that $a_3=672$, and using the fact that $b_3=\alpha b_2-b_1$, we conclude that $\alpha=225$. Clearly, $a_1=b_1$, $a_2=b_2$, and $a_3=b_3$. We claim that $a_n=b_n$ for all $n\ge 1$. We proceed by induction. If $a_k=b_k$ for all $k\le n$, then clearly, \[b_nb_{n-2}-b_{n-1}^2=a_na_{n-2}-a_{n-1}^2=2007.\] We also know by the definition of $b_{n+1}$ that


We know that the RHS is $2007$ by previous work. Therefore, $b_{n+1}b_{n-1}-b_n^2=2007$. After substuting in the values we know, this becomes $b_{n+1}a_{n-1}-a_n^2=2007$. Thinking of this as a linear equation in the variable $b_{n+1}$, we already know that this has the solution $b_{n+1}=a_{n+1}$. Therefore, by induction, $a_n=b_n$ for all $n\ge 1$. We conclude that $a_n$ satisfies the linear recurrence $a_{n+1}=225a_n-a_{n-1}$.

It's easy to prove that $a_n$ is a strictly increasing sequence of integers for $n\ge 3$. Now




The sequence certainly grows fast enough such that $\frac{2007}{a_{2005}a_{2006}}<1$. Therefore, the largest integer less than or equal to this value is $224$.

See also

2007 AIME I (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS