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anonymous
 5 years ago
\[\sum_{n=1}^{\infty}3(2/3)^(n+1)\]
anonymous
 5 years ago
\[\sum_{n=1}^{\infty}3(2/3)^(n+1)\]

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1that was a good effort :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1gonna have to prolly rewrite it tho

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this \[\sum3^{\frac{2}{3}}(n+1)\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no probably not because that would be infinite

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://gracecatholicschool.org/AngelThemes/MML/2010/10/19/151250_726268.gif

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahhhhk a geometric series! ignore the first 3 or rather pull it out front of the summation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually i think it is 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty} (\frac{2}{3})^{n+1}\]=

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0change n + 1 to n and start at n = 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1.... i change my vote to 'im an idiot' :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is 4 if you just want the answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually satellite73, if you change n+1 to n you would have to start from n=2
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