2008 AIME II Problem 6

Attempt Problem 6 of the 2008 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AIME II solutions, or check the answer key.

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6.

The sequence

\{a_n\}

is defined by

\begin{aligned} a_0 &= 1, \\ a_1 &= 1, \\ a_n &= a_{n-1} + \frac{a_{n-1}^2}{a_{n-2}} \quad (n \ge 2). \end{aligned}

The sequence

\{b_n\}

is defined by

\begin{aligned} b_0 &= 1, \\ b_1 &= 3, \\ b_n &= b_{n-1} + \frac{b_{n-1}^2}{b_{n-2}} \quad (n \ge 2). \end{aligned}

Find

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

Answer

Answer: 561

Concepts:recursion factorial telescoping

Difficulty rating: 2460

Small hint:

Divide the recurrence by

a_{n-1}

to see that the ratio

\frac{a_n}{a_{n-1}}

increases by exactly

1

at each step

Big hint:

The ratios give

a_n = n!

and

b_n = \frac{(n+2)!}{2};

form the quotient

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{\frac{34!}{2}}{32!} = \frac{34 \cdot 33}{2} = 561.