Difference between revisions of "1999 AHSME Problems/Problem 13"
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==Problem== | ==Problem== | ||
− | Define a sequence of real numbers <math> a_1</math>, <math> a_2</math>, <math> a_3</math>, <math> \dots</math> by <math> a_1 = 1</math> and <math> a_{n | + | Define a sequence of real numbers <math> a_1</math>, <math> a_2</math>, <math> a_3</math>, <math> \dots</math> by <math> a_1 = 1</math> and <math> a_{n + 1}^3 = 99a_n^3</math> for all <math> n \geq 1</math>. Then <math> a_{100}</math> equals |
<math> \textbf{(A)}\ 33^{33} \qquad | <math> \textbf{(A)}\ 33^{33} \qquad | ||
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==Solution== | ==Solution== | ||
− | {{ | + | We rearrange to get <math>\dfrac{a_{n+1}}{a_n} = \sqrt[3]{99}</math>. Thus we get <math>\dfrac{a_{n+1}}{a_n} = \sqrt[3]{99}</math>, <math>\dfrac{a_{n}}{a_{n-1}} = \sqrt[3]{99}</math>, and so on. Multiplying them all gives <math>\dfrac{a_{n+1}}{a_1} = (\sqrt[3]{99})^{n}</math>. Plugging in <math>n = 99</math> and <math>a_1 = 1</math>, <math>a_{100} = (\sqrt[3]{99})^{99} = 99^{33}</math>, so the answer is <math>\textbf{(C)}</math>. |
==See Also== | ==See Also== | ||
{{AHSME box|year=1999|num-b=12|num-a=14}} | {{AHSME box|year=1999|num-b=12|num-a=14}} |
Revision as of 20:00, 2 June 2011
Problem
Define a sequence of real numbers , , , by and for all . Then equals
Solution
We rearrange to get . Thus we get , , and so on. Multiplying them all gives . Plugging in and , , so the answer is .
See Also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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All AHSME Problems and Solutions |