Difference between revisions of "2009 AMC 12A Problems/Problem 25"
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Our answer is <math>|a_{2009}| = \boxed{\textbf{(A)}\ 0}</math>. | Our answer is <math>|a_{2009}| = \boxed{\textbf{(A)}\ 0}</math>. | ||
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==Note== | ==Note== |
Revision as of 13:57, 11 January 2017
Contents
Problem
The first two terms of a sequence are and
. For
,
![\[a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.\]](http://latex.artofproblemsolving.com/d/f/6/df68c0d3181ac4b7e48b2b812c37cd27f7efaf30.png)
What is ?
Solution 1
Consider another sequence such that
, and
.
The given recurrence becomes
![\begin{align*} a_{n + 2} & = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}} \\ \tan{\theta_{n + 2}} & = \frac {\tan{\theta_n} + \tan{\theta_{n + 1}}}{1 - \tan{\theta_n}\tan{\theta_{n + 1}}} \\ \tan{\theta_{n + 2}} & = \tan(\theta_{n + 1} + \theta_n) \end{align*}](http://latex.artofproblemsolving.com/d/b/a/dba24d7ce865f07f6f06ea651e5b966734f2ea11.png)
It follows that . Since
, all terms in the sequence
will be a multiple of
.
Now consider another sequence such that
, and
. The sequence
satisfies
.
As the number of possible consecutive two terms is finite, we know that the sequence is periodic. Write out the first few terms of the sequence until it starts to repeat.
Note that and
. Thus
has a period of
:
.
It follows that and
. Thus
Our answer is .
Note
It is not actually difficult to list out the terms until it repeats. You will find that the period is 7 starting from term 2.
See also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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