1999 AHSME Problems/Problem 13: Difference between revisions

Latest revision as of 13:34, 5 July 2013

Problem

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

$\textbf{(A)}\ 33^{33} \qquad \textbf{(B)}\ 33^{99} \qquad \textbf{(C)}\ 99^{33} \qquad \textbf{(D)}\ 99^{99} \qquad \textbf{(E)}\ \text{none of these}$

Solution

We rearrange to get $\dfrac{a_{n+1}}{a_n} = \sqrt[3]{99}$ . Thus we get $\dfrac{a_{n+1}}{a_n} = \sqrt[3]{99}$ , $\dfrac{a_{n}}{a_{n-1}} = \sqrt[3]{99}$ , and so on. Multiplying them all gives $\dfrac{a_{n+1}}{a_1} = (\sqrt[3]{99})^{n}$ . Plugging in $n = 99$ and $a_1 = 1$ , $a_{100} = (\sqrt[3]{99})^{99} = 99^{33}$ , so the answer is $\textbf{(C)}$ .

1999 AHSME (Problems • Answer Key • Resources)
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1999 AHSME Problems/Problem 13: Difference between revisions

Latest revision as of 13:34, 5 July 2013

Problem

Solution

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