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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>.
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From this, we can determine that the sequence <math>\frac {b_n}{a_n}</math> corresponds to the [[triangular number]]s.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 22:34, 4 July 2013

Problem

The sequence $\{a_n\}$ is defined by \[a_0 = 1,a_1 = 1, \text{ and } a_n = a_{n - 1} + \frac {a_{n - 1}^2}{a_{n - 2}}\text{ for }n\ge2.\] The sequence $\{b_n\}$ is defined by \[b_0 = 1,b_1 = 3, \text{ and } b_n = b_{n - 1} + \frac {b_{n - 1}^2}{b_{n - 2}}\text{ for }n\ge2.\] Find $\frac {b_{32}}{a_{32}}$.

Solution

Rearranging the definitions, we have \[\frac{a_n}{a_{n-1}} = \frac{a_{n-1}}{a_{n-2}} + 1,\quad \frac{b_n}{b_{n-1}} = \frac{b_{n-1}}{b_{n-2}} + 1\] from which it follows that $\frac{a_n}{a_{n-1}} = 1+ \frac{a_{n-1}}{a_{n-2}} = \cdots = (n-1) + \frac{a_{1}}{a_0} = n$ and $\frac{b_n}{b_{n-1}} = (n-1) + \frac{b_{1}}{b_0} = n+2$. These recursions, $a_{n} = na_{n-1}$ and $b_{n} = (n+2)b_{n-1}$, respectively, correspond to the explicit functions $a_n = n!$ and $b_n = \frac{(n+2)!}{2}$ (after applying our initial conditions). It follows that $\frac{b_{32}}{a_{32}} = \frac{\frac{34!}{2}}{32!} = \frac{34 \cdot 33}{2} = \boxed{561}$.

From this, we can determine that the sequence $\frac {b_n}{a_n}$ 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

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