Difference between revisions of "2005 AIME II Problems/Problem 7"
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== Problem == | == Problem == | ||
− | Let <math> x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}. </math> Find <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> | + | <math>{} (\sqrt[2^n]{5} + 1)(\sqrt[2^n]{5} - 1) = (\sqrt[2^n]{5})^2 - 1^2 = \sqrt[2^{n-1}]{5} - 1 </math>. |
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− | It now becomes apparent that if we multiply the [[numerator]] and [[denominator]] of <math> | + | It now becomes apparent that if we multiply the [[numerator]] and [[denominator]] of <math>\frac{4}{ (\sqrt{5}+1) (\sqrt[4]{5}+1) (\sqrt[8]{5}+1) (\sqrt[16]{5}+1) } </math> by <math>(\sqrt[16]{5} - 1) </math>, the denominator will [[telescope]] to <math>\sqrt[1]{5} - 1 = 4 </math>, so |
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− | <math> | + | <math>x = \frac{4(\sqrt[16]{5} - 1)}{4} = \sqrt[16]{5} - 1</math>. |
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 22:22, 4 July 2013
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 .
See Also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.