Difference between revisions of "2008 AIME II Problems/Problem 6"
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Latest revision as of 23:34, 4 July 2013
Problem
The sequence 
is defined by
![\[a_0 = 1,a_1 = 1, \text{ and } a_n = a_{n - 1} + \frac {a_{n - 1}^2}{a_{n - 2}}\text{ for }n\ge2.\]](http://latex.artofproblemsolving.com/4/0/8/408d131150595befc5ae2662573b99ec545e5819.png)
The sequence 
is defined by
![\[b_0 = 1,b_1 = 3, \text{ and } b_n = b_{n - 1} + \frac {b_{n - 1}^2}{b_{n - 2}}\text{ for }n\ge2.\]](http://latex.artofproblemsolving.com/e/3/a/e3a90aa84fa76712b0de3ce0ce81cdef9c1d3e8d.png)
Find 
.
Solution
Rearranging the definitions, we have
![\[\frac{a_n}{a_{n-1}} = \frac{a_{n-1}}{a_{n-2}} + 1,\quad \frac{b_n}{b_{n-1}} = \frac{b_{n-1}}{b_{n-2}} + 1\]](http://latex.artofproblemsolving.com/e/a/c/eac923f11fcde9dfa8ebb1b40688b782b7f3c02b.png)
from which it follows that 
and 
. These recursions, 
and 
, respectively, correspond to the explicit functions 
and 
(after applying our initial conditions). It follows that 
.
From this, we can determine that the sequence 
corresponds to the triangular numbers.
See also
| 2008 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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. 
