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Difference between revisions of "1971 AHSME Problems/Problem 7"
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==Problem==
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<math>2^{-(2k+1)}-2^{-(2k-1)}+2^{-2k}</math> is equal to
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<math>\textbf{(A) }2^{-2k}\qquad \textbf{(B) }2^{-(2k-1)}\qquad \textbf{(C) }-2^{-(2k+1)}\qquad \textbf{(D) }0\qquad  \textbf{(E) }2</math>
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==Solution==
 
<math>Let\ x\ equal\ 2^{-2k}\ \\*
 
<math>Let\ x\ equal\ 2^{-2k}\ \\*
 
From\ this\ we\ get \frac{x}{2}-(\frac{-x}{\frac{-1}{2}})+x\ by\ using\ power\ rule.\  
 
From\ this\ we\ get \frac{x}{2}-(\frac{-x}{\frac{-1}{2}})+x\ by\ using\ power\ rule.\  

Latest revision as of 17:13, 23 June 2021

Problem

$2^{-(2k+1)}-2^{-(2k-1)}+2^{-2k}$ is equal to

$\textbf{(A) }2^{-2k}\qquad \textbf{(B) }2^{-(2k-1)}\qquad \textbf{(C) }-2^{-(2k+1)}\qquad \textbf{(D) }0\qquad \textbf{(E) }2$

Solution

$Let\ x\ equal\ 2^{-2k}\ \\* From\ this\ we\ get \frac{x}{2}-(\frac{-x}{\frac{-1}{2}})+x\ by\ using\ power\ rule.\ \\*Now\ we\ can\ see\ this\ simplies\ to\ \frac{-x}{2}\ \\*Looking\ at\ \frac{x}{2} we\ can\ clearly\ see\ that\ -(\frac{x}{2}) is\ equal\ to\ -2^{-(2k+1)}\ \\*Thus\ our\ answer\ is\ c$

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