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@Jhannybean
I have never gotten to series in calculus Lala, sorry.. but I called someone better for you :)
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 \[\large \sum_{n-1}^{\infty}-4(-\frac{1}{2})^{n-1} = C\]
thank you @Luigi0210
Because it converges, your sum : \[\large S=\frac{a}{1-r}\] where \[\large r=-\frac{1}{2}\]
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that stuff looks so familiar >.<
TRue that
so that is the answer?
yes
oh ok!
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