1971 AHSME Problems/Problem 7

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: $\begin{aligned} 2^{- (2 k + 1)} - 2^{- (2 k - 1)} + 2^{- 2 k} & = 2^{- 2 k - 1} - 2^{- 2 k + 1} + 2^{- 2 k} \\ = \frac{2^{- 2 k}}{2} - 2 \cdot 2^{- 2 k} + 2^{- 2 k} \\ = 2^{- 2 k} (\frac{1}{2} - 2 + 1) \\ = 2^{- 2 k} (- \frac{1}{2}) \\ = - \frac{2^{- 2 k}}{2} \\ = - 2^{- 2 k - 1} \\ = (C) - 2^{- (2 k + 1)} . \end{aligned}$

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
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1971 AHSME Problems/Problem 7

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