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2008 AIME II Problems/Problem 6

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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 • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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