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does \[\frac{1}{1-2x}=2^n\]?
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not if x = 3!
what are the steps to get it from the left to the right?
apparently discrete math is my achilles heel lol
this question honestly does not make any sense. you have an x on one side and n on the other.
^^I agree.
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hmmm..... well, i think i write it right..
should there be an 'n' up top instead of a 1?
Then I think they are not equal.
ooooooooooooohhhhhhhhh i bet i see what this should be. maybe. \[\frac{1}{1-x}=\Sigma x^n\] for -1<x<1 now if we replace x by 2x we get \[\frac{1}{1-2x}=\Sigma (2x)^n = \Sigma 2^nx^n\] for \[-\frac{1}{2}<x<\frac{1}{2}\] perhaps this is what you had in mind?
that looks more like it :)
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