Difference between revisions of "2008 AIME II Problems/Problem 6"
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<cmath>\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</cmath> | <cmath>\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</cmath> | ||
from which it follows that <math>\frac{a_n}{a_{n-1}} = 1+ \frac{a_{n-1}}{a_{n-2}} = \cdots = (n-1) + \frac{a_{1}}{a_0} = n</math> and <math>\frac{b_n}{b_{n-1}} = (n-1) + \frac{b_{1}}{b_0} = n+2</math>. These [[recursion]]s, <math>a_{n} = na_{n-1}</math> and <math>b_{n} = (n+2)b_{n-1}</math>, respectively, correspond to the explicit functions <math>a_n = n!</math> and <math>b_n = \frac{(n+2)!}{2}</math> (after applying our initial conditions). It follows that <math>\frac{b_{32}}{a_{32}} = \frac{\frac{34!}{2}}{32!} = \frac{34 \cdot 33}{2} = \boxed{561}</math>. | from which it follows that <math>\frac{a_n}{a_{n-1}} = 1+ \frac{a_{n-1}}{a_{n-2}} = \cdots = (n-1) + \frac{a_{1}}{a_0} = n</math> and <math>\frac{b_n}{b_{n-1}} = (n-1) + \frac{b_{1}}{b_0} = n+2</math>. These [[recursion]]s, <math>a_{n} = na_{n-1}</math> and <math>b_{n} = (n+2)b_{n-1}</math>, respectively, correspond to the explicit functions <math>a_n = n!</math> and <math>b_n = \frac{(n+2)!}{2}</math> (after applying our initial conditions). It follows that <math>\frac{b_{32}}{a_{32}} = \frac{\frac{34!}{2}}{32!} = \frac{34 \cdot 33}{2} = \boxed{561}</math>. | ||
| + | |||
| + | == Solution 2 == | ||
| + | It's very easy just to write out the first couple terms of each and find a pattern. The terms of <math>{a_n}</math> are 1,1,2,6,and 24. The terms of <math>{b_n}</math> are 1,3,12,60,and 360. When we divide <math>{b_n}</math> by <math>{a_n}</math>, we get 1,3,6,10, and 15, the triangular numbers! But since 1 corresponds to <math>{a_0}/{b_0}</math>, the answer is the 33rd triangular number, which is 561. | ||
== See also == | == See also == | ||
Revision as of 18:10, 4 April 2008
Contents
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 
.
Solution 2
It's very easy just to write out the first couple terms of each and find a pattern. The terms of 
are 1,1,2,6,and 24. The terms of 
are 1,3,12,60,and 360. When we divide by , we get 1,3,6,10, and 15, the triangular numbers! But since 1 corresponds to 
, the answer is the 33rd triangular number, which is 561.
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 | ||
