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anonymous

  • 5 years ago

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

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  1. amistre64
    • 5 years ago
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    that was a good effort :)

  2. amistre64
    • 5 years ago
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    gonna have to prolly rewrite it tho

  3. anonymous
    • 5 years ago
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    is this \[\sum3^{\frac{2}{3}}(n+1)\]?

  4. anonymous
    • 5 years ago
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    no probably not because that would be infinite

  5. anonymous
    • 5 years ago
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    http://gracecatholicschool.org/AngelThemes/MML/2010/10/19/151250_726268.gif

  6. anonymous
    • 5 years ago
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    ahhhhk a geometric series! ignore the first 3 or rather pull it out front of the summation

  7. anonymous
    • 5 years ago
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    6

  8. anonymous
    • 5 years ago
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    just 6?

  9. anonymous
    • 5 years ago
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    yeah

  10. anonymous
    • 5 years ago
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    actually i think it is 4

  11. anonymous
    • 5 years ago
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    soooo 4?

  12. anonymous
    • 5 years ago
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    sorry 4...lol

  13. anonymous
    • 5 years ago
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    0 or 4?

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

  15. anonymous
    • 5 years ago
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    4

  16. anonymous
    • 5 years ago
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    change n + 1 to n and start at n = 0

  17. amistre64
    • 5 years ago
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    .... i change my vote to 'im an idiot' :)

  18. anonymous
    • 5 years ago
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    ok so 4 it is

  19. anonymous
    • 5 years ago
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    it is 4 if you just want the answer

  20. anonymous
    • 5 years ago
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    yesssss

  21. anonymous
    • 5 years ago
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    thnx

  22. anonymous
    • 5 years ago
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    NO THANKK YOUUUU :)

  23. anonymous
    • 5 years ago
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    actually satellite73, if you change n+1 to n you would have to start from n=2

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