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Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.2I have never gotten to series in calculus Lala, sorry.. but I called someone better for you :)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.3In 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 \[\large \sum_{n1}^{\infty}4(\frac{1}{2})^{n1} = C\]

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

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.2that stuff looks so familiar >.<

lala2
 one year ago
Best ResponseYou've already chosen the best response.0you think you could help with another?
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