Difference between revisions of "1971 AHSME Problems/Problem 7"
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+ | ==Problem== | ||
+ | <math>2^{-(2k+1)}-2^{-(2k-1)}+2^{-2k}</math> is equal to | ||
+ | |||
+ | <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> | ||
+ | ==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.\ |
Revision as of 16:13, 23 June 2021
Problem
is equal to
Solution