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Luigi0210Best ResponseYou've already chosen the best response.2
I have never gotten to series in calculus Lala, sorry.. but I called someone better for you :)
 10 months ago

JhannybeanBest ResponseYou've already chosen the best response.3
In a geometric series, the series will converge if \[\large r<1\]In this case, \[\frac{1}{2} <1 \implies \frac{1}{2}< 1\] therefore the \[\large \sum_{n1}^{\infty}4(\frac{1}{2})^{n1} = C\]
 10 months ago

JhannybeanBest ResponseYou've already chosen the best response.3
Because it converges, your sum : \[\large S=\frac{a}{1r}\] where \[\large r=\frac{1}{2}\]
 10 months ago

Luigi0210Best ResponseYou've already chosen the best response.2
that stuff looks so familiar >.<
 10 months ago

lala2Best ResponseYou've already chosen the best response.0
you think you could help with another?
 10 months ago
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