Define a sequence of real numbers a_1, a_2, a_3, … by a_1 = 1 and a_n+1^3 = 99 a_n^3 for all n ≥ 1. Then a_100 equals

1999 AMC 12 Problem 13 — Solution

1999 AMC 12 Problem 13

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13.

Define a sequence of real numbers $a_1, a_2, a_3, \ldots$ by $a_1 = 1$ and $a_{n+1}^3 = 99 a_n^3$ for all $n \ge 1.$ Then $a_{100}$ equals

$33^{33}$

$33^{99}$

$99^{33}$

$99^{99}$

none of these

Answer: C

Concepts:geometric sequence exponent

Difficulty rating: 1420

Solution:

Taking cube roots, $a_{n+1} = \sqrt[3]{99}\, a_n,$ so the sequence is geometric with first term $1$ and ratio $\sqrt[3]{99}.$ Then $a_{100} = \left(\sqrt[3]{99}\right)^{99} = 99^{33}.$

Thus, the correct answer is C.

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