# Difference between revisions of "2005 AIME II Problems/Problem 7"

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== Problem == | == Problem == | ||

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Let <math> x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}. </math> Find <math> (x+1)^{48}. </math> | Let <math> x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}. </math> Find <math> (x+1)^{48}. </math> | ||

== Solution == | == Solution == | ||

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− | <math> | + | We note that in general, |

− | <math> | + | |

− | + | <center> | |

+ | <math>\displaystyle {} (\sqrt[2^n]{5} + 1)(\sqrt[2^n]{5} - 1) = (\sqrt[2^n]{5})^2 - 1^2 = \sqrt[2^{n-1}]{5} - 1 </math>. | ||

+ | </center> | ||

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+ | It now becomes apparent that if we multiply the [[numerator]] and [[denominator]] of <math> \displaystyle \frac{4}{ (\sqrt{5}+1) (\sqrt[4]{5}+1) (\sqrt[8]{5}+1) (\sqrt[16]{5}+1) } </math> by <math>\displaystyle (\sqrt[16]{5} - 1) </math>, the denominator will [[telescope]] to <math> \displaystyle \sqrt[1]{5} - 1 = 4 </math>, so | ||

+ | |||

+ | <center> | ||

+ | <math> \displaystyle x = \frac{4(\sqrt[16]{5} - 1)}{4} = \sqrt[16]{5} - 1</math>. | ||

+ | </center> | ||

+ | |||

+ | It follows that <math>(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = 125</math> | ||

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== See Also == | == See Also == | ||

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*[[2005 AIME II Problems]] | *[[2005 AIME II Problems]] | ||

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+ | [[Category:Introductory Algebra Problems]] |

## Revision as of 16:49, 7 September 2006

## Problem

Let Find

## Solution

We note that in general,

.

It now becomes apparent that if we multiply the numerator and denominator of by , the denominator will telescope to , so

.

It follows that