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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    You 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...

  2. anonymous
    • 5 years ago
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    \[\sum_{k=0}^{\infty}\frac{3^{k-1}}{4^{3k+1}}=\sum_{k=0}^{\infty}\frac{3^k.3^{-1}}{(4^3)^k.4}=\frac{1}{12}\sum_{k=0}^{k=\infty}(\frac{3}{64})^k\]

  3. anonymous
    • 5 years ago
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    The geometric series has first term a=1 and common ratio r=3/64. The sum is then,\[\frac{a}{1-r}=\frac{1}{1-3/64}=\frac{64}{61}\]

  4. anonymous
    • 5 years ago
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    You have to now multiply this answer by 1/12 to get your sum:\[\sum_{k=0}^{\infty}\frac{3^{k-1}}{4^{3k+1}}=\frac{16}{183}\]

  5. anonymous
    • 5 years ago
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    Please fan me :) I punch out answers and no one bothers to reward! Cheers.

  6. anonymous
    • 5 years ago
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    Wow thank you and how do I fan you? I am new to this site

  7. anonymous
    • 5 years ago
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    There 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...

  8. anonymous
    • 5 years ago
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    oh lol I had refresh it to see the 'become a fan' button but anyway thank you for your help!

  9. anonymous
    • 5 years ago
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    cheers...no worries.

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