Difference between revisions of "2007 AIME I Problems/Problem 14"
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Find the [[floor function|greatest integer]] that does not exceed <math>\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}</math> | Find the [[floor function|greatest integer]] that does not exceed <math>\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}</math> | ||
− | == Solution 1== | + | == Solution == |
+ | |||
+ | === Solution 1 === | ||
We are given that | We are given that | ||
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<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>. | <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>. | ||
− | == Solution 2== | + | === Solution 2=== |
The equation <math>a_{n+1}a_{n-1}-a_n^2=2007</math> looks like the determinant <cmath>\left|\begin{array}{cc}a_{n+1}&a_n\\a_n&a_{n-1}\end{array}\right|=2007.</cmath> | The equation <math>a_{n+1}a_{n-1}-a_n^2=2007</math> looks like the determinant <cmath>\left|\begin{array}{cc}a_{n+1}&a_n\\a_n&a_{n-1}\end{array}\right|=2007.</cmath> | ||
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 <math>b_n</math> defined by <math>b_1=b_2=3</math> and <math>b_{n+1}=\alpha b_n+\beta b_{n-1}</math> for <math>n\ge 2</math>. We wish to find <math>\alpha</math> and <math>\beta</math> such that <math>a_n=b_n</math> for all <math>n\ge 1</math>. To do this, we use the following matrix form of a linear recurrence relation | 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 <math>b_n</math> defined by <math>b_1=b_2=3</math> and <math>b_{n+1}=\alpha b_n+\beta b_{n-1}</math> for <math>n\ge 2</math>. We wish to find <math>\alpha</math> and <math>\beta</math> such that <math>a_n=b_n</math> for all <math>n\ge 1</math>. To do this, we use the following matrix form of a linear recurrence relation |
Revision as of 19:41, 25 January 2014
Problem
A sequence is defined over non-negative integral indexes in the following way: ,
.
Find the greatest integer that does not exceed
Solution
Solution 1
We are given that
,
.
Add these two equations to get
.
This is an invariant. Defining for each
, the above equation means
.
We can thus calculate that . Now notice that
. This means that
. It is only a tiny bit less because all the
are greater than
, so we conclude that the floor of
is
.
Solution 2
The equation looks like the determinant
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
defined by
and
for
. We wish to find
and
such that
for all
. To do this, we use the following matrix form of a linear recurrence relation
When we take determinants, this equation becomes
We want for all
. Therefore, we replace the two matrices by
to find that
Therefore,
. Computing that
, and using the fact that
, we conclude that
. Clearly,
,
, and
. We claim that
for all
. We proceed by induction. If
for all
, then clearly,
We also know by the definition of
that
We know that the RHS is by previous work. Therefore,
. After substuting in the values we know, this becomes
. Thinking of this as a linear equation in the variable
, we already know that this has the solution
. Therefore, by induction,
for all
. We conclude that
satisfies the linear recurrence
.
It's easy to prove that is a strictly increasing sequence of integers for
. Now
The sequence certainly grows fast enough such that . Therefore, the largest integer less than or equal to this value is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.