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.