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

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

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

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