Question

The sequence $\{a_n\}$ is defined by $a_0 = 1, \quad a_1 = 1, \quad \text{and} \quad a_n = a_{n-1} + \frac{a_{n-1}^2}{a_{n-2}} \quad \text{for } n \ge 2.$

The sequence $\{b_n\}$ is defined by $b_0 = 1, \quad b_1 = 3, \quad \text{and} \quad b_n = b_{n-1} + \frac{b_{n-1}^2}{b_{n-2}} \quad \text{for } n \ge 2.$

Find $\frac{b_{32}}{a_{32}}.$

答案

Solution:

Dividing the recurrence by $a_{n-1}$ gives $\frac{a_n}{a_{n-1}} = 1 + \frac{a_{n-1}}{a_{n-2}},$ so the consecutive-term ratio increases by exactly $1$ each step. For $\{a_n\}$ the first ratio is $\frac{a_1}{a_0} = 1,$ so $\frac{a_n}{a_{n-1}} = n$ and $a_n = n!.$ The same computation applies to $\{b_n\},$ whose first ratio is $\frac{b_1}{b_0} = 3,$ so $\frac{b_n}{b_{n-1}} = n + 2$ and $b_n = \frac{(n+2)!}{2}.$

Therefore $\frac{b_{32}}{a_{32}} = \frac{34!/2}{32!} = \frac{34 \cdot 33}{2} = 561.$

2008 AIME II Problem 6

Solution:

Problem 6 in Other Years