Difference between revisions of "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 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

$\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).$

When we take determinants, this equation becomes

$\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).$

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

$\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).$

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

$\frac{a_{2007}^2+a_{2006}^2}{a_{2007}a_{2006}}=\frac{a_{2007}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}=\frac{225a_{2006}-a_{2005}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}.$

$=225+\frac{a_{2006}}{a_{2007}}-\frac{a_{2005}}{a_{2006}}=225+\frac{a_{2006}^2-a_{2005}a_{2007}}{a_{2005}a_{2006}}.$

$=225-\frac{2007}{a_{2005}a_{2006}}.$

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 $224$ .

Solution 3 ( generalized )

This is a more elementary and rigorous solution to a slightly generalized version. The defining recursive sequence is generalized to

$a_{n+1}a_{n-1} = a_n^2 + 9k, ---------(1)$ where $k$ is a positive integer and $a_0 = a_1 = 3.$

Lemma 1 : For $n \geq 1$ , $a_{n+1} = ( k + 2)a_n - a_{n-1}. ---------(2)$ We shall prove by induction. From (1), $a_2 = 3k + 3$ . From the lemma, $a_2 = (k + 2) 3 - 3 = 3k + 3.$ Base case proven. Assume that the lemma is true for some $t \geq 1$ . Then, eliminating the $a_{t-1}$ using (1) and (2) gives

$(k+2)a_ta_{t+1} = a_t^2 + a_{t+1}^2 + 9k. ---------(3)$

It follows from (2) that

$(k+2)a_{t+1} - a_t =\frac{(k+2)a_{t+1}a_t - a_t^2}{a_t} =\frac{a_{t+1}^2 + 9k}{a_t} =a_{t+2},$

where the last line followed from (1) for case $n = t+1$ .

Lemma 2 : For $n \geq 0,$ $a_{n+1} \geq a_{n}.$ Base case is obvious. Assume that $a_{t+1} \geq a_{t}$ for some $t \geq 0$ . Then it follows that $a_{t+2} =\frac{a_{t+1}^2 + 9k}{a_t} = a_{t+1}(\frac{a_{t+1}}{a_t} ) + 9k \geq a_{t+1} + 9k > a_{t+1}.$

This completes the induction.

Lemma 3 : For $n \geq 1,$ $a_n a_{n+1} > 9k$

Using (1) and Lemma 2, for $n \geq 1,$ $a_{n+1}a_n \geq a_{n+1}a_{n-1} = a_n^2 + 9k > 9k$

Finally, using (3), for $n \geq 1,$ $\frac{a_n^2 + a_{n+1}^2}{a_n a_{n+1}} =\frac{(k+2)a_n a_{n+1} - 9k}{a_n a_{n+1}} = k+2 -\frac{9k}{a_n a_{n+1}}.$ Using lemma 3, the largest integer less than or equal to this value would be $k + 1$ .

@@ Line 53: / Line 53: @@
 The sequence certainly grows fast enough such that <math>\frac{2007}{a_{2005}a_{2006}}<1</math>. Therefore, the largest integer less than or equal to this value is <math>224</math>.
+===Solution 3 ( generalized )===
+This is a more elementary and rigorous solution to a slightly generalized version. The defining recursive sequence is generalized to
+<cmath>
+a_{n+1}a_{n-1} = a_n^2 + 9k, ---------(1)
+</cmath>
+where <math> k </math> is a positive integer and <math> a_0 = a_1 = 3. </math>
+Lemma 1  : For <math>n \geq 1</math>,
+<cmath>
+ a_{n+1} = ( k + 2)a_n - a_{n-1}. ---------(2)
+</cmath>
+We shall prove by induction. From (1), <math> a_2 = 3k + 3 </math>. From the lemma, <math>a_2 = (k + 2) 3 - 3 = 3k + 3.</math> Base case proven. Assume that the lemma is true for some <math> t \geq 1 </math>. Then, eliminating the <math>a_{t-1}</math> using (1) and (2) gives
+<cmath>
+ (k+2)a_ta_{t+1} = a_t^2 + a_{t+1}^2 + 9k. ---------(3)
+</cmath>
+It follows from (2) that
+<cmath>
+(k+2)a_{t+1} - a_t  =\frac{(k+2)a_{t+1}a_t - a_t^2}{a_t}  =\frac{a_{t+1}^2 + 9k}{a_t} =a_{t+2},
+</cmath>
+where the last line followed from (1) for case <math> n = t+1 </math>.
+Lemma 2 : For <math>n \geq 0,</math>
+<cmath>
+a_{n+1} \geq a_{n}.
+</cmath>
+Base case is obvious. Assume that <math>a_{t+1} \geq a_{t}</math> for some <math>t \geq 0</math>. Then it follows that
+<cmath>
+a_{t+2} =\frac{a_{t+1}^2 + 9k}{a_t}  = a_{t+1}(\frac{a_{t+1}}{a_t} ) + 9k \geq a_{t+1} + 9k  > a_{t+1}.
+</cmath>
+This completes the induction.
+Lemma 3 : For <math>n \geq 1,</math>
+<cmath>
+a_n a_{n+1} > 9k
+</cmath>
+Using (1) and Lemma 2, for <math>n \geq 1,</math>
+<cmath>
+a_{n+1}a_n \geq a_{n+1}a_{n-1}  = a_n^2 + 9k  > 9k
+</cmath>
+Finally, using (3), for <math>n \geq 1, </math>
+<cmath>
+\frac{a_n^2 + a_{n+1}^2}{a_n a_{n+1}} =\frac{(k+2)a_n a_{n+1} - 9k}{a_n a_{n+1}} = k+2 -\frac{9k}{a_n a_{n+1}}.
+</cmath>
+Using lemma 3, the largest integer less than or equal to this value would be <math>k + 1</math>.
 == 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
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