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Difference between revisions of "1971 AHSME Problems/Problem 7"
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<math>Let\ x\ equal\ 2^{-2k}\ \\*
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==Problem==
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From\ this\ we\ get \frac{x}{2}-(\frac{-x}{\frac{-1}{2}})+x\ by\ using\ power\ rule.\  
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<math>2^{-(2k+1)}-2^{-(2k-1)}+2^{-2k}</math> is equal to
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\\*Now\ we\ can\ see\ this\ simplies\ to\ \frac{-x}{2}\
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\\*Looking\ at\ \frac{x}{2} we\ can\ clearly\ see\ that\ \frac{-x}{2} is\ equal\ to\ -2^{-(2k+1)}\
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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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\\*Thus\ our\ answer\ is\ c</math>
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==Solution==
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By using the properties of [[exponentiation#Basic Properties|exponents]], we can simplify the given expression as follows to obtain our answer:
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\begin{align*}
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2^{-(2k+1)} - 2^{-(2k-1)} + 2^{-2k} &= 2^{-2k-1} - 2^{-2k+1} + 2^{-2k} \\
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&= \frac{2^{-2k}}2 - 2\cdot2^{-2k} + 2^{-2k} \\
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&= 2^{-2k}(\frac12 - 2 + 1) \\
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&= 2^{-2k}(-\frac12) \\
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&= -\frac{2^{-2k}}2 \\
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&= -2^{-2k-1} \\
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&= \boxed{\textbf{(C) }-2^{-(2k+1)}}.
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\end{align*}
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== See Also ==
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{{AHSME 35p box|year=1971|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 11:07, 1 August 2024

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

By using the properties of exponents, we can simplify the given expression as follows to obtain our answer: \begin{align*} 2^{-(2k+1)} - 2^{-(2k-1)} + 2^{-2k} &= 2^{-2k-1} - 2^{-2k+1} + 2^{-2k} \\ &= \frac{2^{-2k}}2 - 2\cdot2^{-2k} + 2^{-2k} \\ &= 2^{-2k}(\frac12 - 2 + 1) \\ &= 2^{-2k}(-\frac12) \\ &= -\frac{2^{-2k}}2 \\ &= -2^{-2k-1} \\ &= \boxed{\textbf{(C) }-2^{-(2k+1)}}. \end{align*}

See Also

1971 AHSC (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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