Difference between revisions of "2008 AIME II Problems/Problem 6"
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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>. | ||
− | + | From this, we can determine that the sequence <math>\frac {b_n}{a_n}</math> corresponds to the [[triangular number]]s. | |
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== See also == | == See also == |
Revision as of 15:24, 7 April 2008
Problem
The sequence is defined by The sequence is defined by Find .
Solution
Rearranging the definitions, we have 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 |