Difference between revisions of "1971 AHSME Problems/Problem 7"
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| − | Let\ x\ equal\ 2^{-2k} | + | ==Problem== |
| − | \ | + | <math>2^{-(2k+1)}-2^{-(2k-1)}+2^{-2k}</math> is equal to |
| − | 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}\ | + | <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> |
| − | Looking\ at\ \frac{x}{2} we\ can\ clearly\ see\ that\ \frac{ | + | ==Solution== |
| − | Thus\ our\ answer\ is\ c | + | <math>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</math> | ||
Latest revision as of 17:13, 23 June 2021
Problem
is equal to
Solution
