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

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13.

Define a sequence of real numbers

a_1,

a_2,

a_3,

\ldots

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

Take cube roots:

a_{n+1} = \sqrt[3]{99}\, a_n

This is geometric with ratio

\sqrt[3]{99},

a_{100} = (\sqrt[3]{99})^{99}

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.