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Silenthill
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Find the sum of the Series:
Sigma (1)^(n1) (3/4^n) from n=1 to infinity
 2 years ago
 2 years ago
Silenthill Group Title
Find the sum of the Series: Sigma (1)^(n1) (3/4^n) from n=1 to infinity
 2 years ago
 2 years ago

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blockcolder Group TitleBest ResponseYou've already chosen the best response.6
You mean this: \[\sum_{n=1}^{\infty}(1)^{n1}\frac{3}{4^n}\]
 2 years ago

blockcolder Group TitleBest ResponseYou've already chosen the best response.6
This is an alternating geometric series with r=1/4 and a=3/4. Thus, the sum is \[\frac{\frac{3}{4}}{1+\frac{1}{4}}=\frac{3}{5}\].
 2 years ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
the common ratio is 1/4 and 1st term a = 3/4 then \[s _{\infty} = a/(1r)\]
 2 years ago

Silenthill Group TitleBest ResponseYou've already chosen the best response.0
Thank you!
 2 years ago

blockcolder Group TitleBest ResponseYou've already chosen the best response.6
No problem. :D
 2 years ago
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