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anonymous
 5 years ago
How do you find the sum of the series of 3^(k1)/4^(3k+1) ; k = 0 to infinity? (infinite series)
anonymous
 5 years ago
How do you find the sum of the series of 3^(k1)/4^(3k+1) ; k = 0 to infinity? (infinite series)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You just have to do some manipulation on the series to see that it's actually geometric, and then all you need to do is use the sum of a geometric series formula. So...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{k=0}^{\infty}\frac{3^{k1}}{4^{3k+1}}=\sum_{k=0}^{\infty}\frac{3^k.3^{1}}{(4^3)^k.4}=\frac{1}{12}\sum_{k=0}^{k=\infty}(\frac{3}{64})^k\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The geometric series has first term a=1 and common ratio r=3/64. The sum is then,\[\frac{a}{1r}=\frac{1}{13/64}=\frac{64}{61}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to now multiply this answer by 1/12 to get your sum:\[\sum_{k=0}^{\infty}\frac{3^{k1}}{4^{3k+1}}=\frac{16}{183}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Please fan me :) I punch out answers and no one bothers to reward! Cheers.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wow thank you and how do I fan you? I am new to this site

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There should be a [thumbs up] icon next to my name and the words, "Become a fan" underlined as a link. You just click on that...it's blue...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh lol I had refresh it to see the 'become a fan' button but anyway thank you for your help!
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