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Difference between revisions of "1971 AHSME Problems/Problem 32"

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== Solution ==
 
== Solution ==
  
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First, we write the expression as <math>(1+\frac{1}{\sqrt[32]{2}})(1+\frac{1}{\sqrt[32]{2}})(1+\frac{1}{\sqrt[32]{2}})(1+\frac{1}{\sqrt[32]{2}})(1+\frac{1}{\sqrt[32]{2}})</math>
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First, we write the expression as <math>(1+\frac{1}{\sqrt[32]{2}})(1+\frac{1}{\sqrt[16]{2}})(1+\frac{1}{\sqrt[8]{2}})(1+\frac{1}{\sqrt[4]{2}})(1+\frac{1}{\sqrt[2]{2}})</math>

Revision as of 21:46, 16 August 2019

Problem 32

If $s=(1+2^{-\frac{1}{32}})(1+2^{-\frac{1}{16}})(1+2^{-\frac{1}{8}})(1+2^{-\frac{1}{4}})(1+2^{-\frac{1}{2}})$, then $s$ is equal to

$\textbf{(A) }\textstyle{\frac{1}{2}}(1-2^{-\frac{1}{32}})^{-1}\qquad \textbf{(B) }(1-2^{-\frac{1}{32}})^{-1}\qquad \textbf{(C) }1-2^{-\frac{1}{32}}\qquad \\ \textbf{(D) }\textstyle{\frac{1}{2}}(1-2^{-\frac{1}{32}})\qquad \textbf{(E) }\frac{1}{2}$

Solution

First, we write the expression as $(1+\frac{1}{\sqrt[32]{2}})(1+\frac{1}{\sqrt[16]{2}})(1+\frac{1}{\sqrt[8]{2}})(1+\frac{1}{\sqrt[4]{2}})(1+\frac{1}{\sqrt[2]{2}})$

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