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Math Help - Series problem (again!)

  1. #1
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    Series problem (again!)

    Hi guys, so i was doing this one question and cant find a way through. It goes by like this :
    What is the condition such that the series

    1+[7/(3x+5)]+[7+(3x+5)^2].... is converge
    hence show that the sum of the above series is 3x-5/3x-12
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  2. #2
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    Re: Series problem (again!)

    Quote Originally Posted by nisbahmumtaz View Post
    1+[7/(3x+5)]+[7+(3x+5)^2].... is converge
    hence show that the sum of the above series is 3x-5/3x-12
    Could you write the series and the answer correctly? I suspect the answer is not 3x-\frac{5}{3x}-12.
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  3. #3
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    Re: Series problem (again!)

    My bad. Sorry. It's (3x-5)/(3x-12)
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  4. #4
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    Re: Series problem (again!)

    Quote Originally Posted by nisbahmumtaz View Post
    Hi guys, so i was doing this one question and cant find a way through. It goes by like this :
    What is the condition such that the series
    1+[7/(3x+5)]+[7+(3x+5)^2].... is converge
    hence show that the sum of the above series is 3x-5/3x-12
    Is that + correct? Could it be \frac{7}{(3x+5)}+\frac{7}{(3x+5)^2}+\frac{7}{(3x+5  )^3}\cdots~?
    If not then it diverges.
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  5. #5
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    Re: Series problem (again!)

    And 1+\sum_{i=1}^\infty\frac{7}{(3x+5)^i}=\frac{3x+11}  {3x+4} (using geometric progression).

    Edit: added ^i to the formula.
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  6. #6
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    Re: Series problem (again!)

    Quote Originally Posted by Plato View Post
    Is that + correct? Could it be \frac{7}{(3x+5)}+\frac{7}{(3x+5)^2}+\frac{7}{(3x+5  )^3}\cdots~?
    If not then it diverges.
    Yikes. Darn it this iphone keyboard. Its supposed to be " / " not "+"
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  7. #7
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    Re: Series problem (again!)

    Quote Originally Posted by nisbahmumtaz View Post
    Yikes. Darn it this iphone keyboard. Its supposed to be " / " not "+"
    The see reply #5 for the correct answer.
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