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> (x+1)^{48} | + | Let <math> x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}. </math> Find <math>\displaystyle (x+1)^{48}</math>. |
== Solution == | == Solution == | ||
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− | It follows that <math>(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = 125</math> | + | It follows that <math>(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = 125</math>. |
== See Also == | == See Also == | ||
− | + | {{AIME box|year=2005|n=II|num-b=6|num-a=8}} | |
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 19:55, 21 March 2007
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 |