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Math Help - Summation

  1. #1
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    Summation

    Express \sum_{n = 1}^N(\frac{1}{8(3^{-2n})}) in terms of N.

     \sum_{n = 1}^N(\frac{1}{8(3^{-2n})}) = (\frac{1}{(8(3^{-2})})+(\frac{1}{(8(3^{-4})})+...+(\frac{1}{(8(3^{-2N})})

    =\frac{1\frac{1}{8}(1-(\frac{1}{9})^N)}{1-\frac{1}{9}}

    =\frac{1\frac{1}{8}(1-(\frac{1}{9})^N)}{\frac{8}{9}}

     =1\frac{17}{64}(1-(\frac{1}{9})^N)

    but ans is \frac{9}{64}(9^N-1)
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  2. #2
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    I would write \displaystyle \sum_{n = 1}^N{\frac{1}{8\left(3^{-2n}\right)}} = \frac{1}{8}\sum_{n = 1}^N{9^n}.

    This is a geometric series with \displaystyle a = 9, r = 9, n = N. Now apply \displaystyle S_n = \frac{a(r^n - 1)}{r - 1}.
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  3. #3
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    Quote Originally Posted by Punch View Post
    Express \sum_{n = 1}^N(\frac{1}{8(3^{-2n})}) in terms of N.

     \sum_{n = 1}^N(\frac{1}{8(3^{-2n})}) = (\frac{1}{(8(3^{-2})})+(\frac{1}{(8(3^{-4})})+...+(\frac{1}{(8(3^{-2N})})

    =\frac{1\frac{1}{8}(1-(\frac{1}{9})^N)}{1-\frac{1}{9}}

    \sum_{n=1}^{N} \frac{1}{8\ 3^{-2n}} = \frac{1}{8}

    =\frac{1\frac{1}{8}(1-(\frac{1}{9})^N)}{\frac{8}{9}}

     =1\frac{17}{64}(1-(\frac{1}{9})^N)

    but ans is \frac{9}{64}(9^N-1)
    \sum_{n=1}^{N} \frac{1}{8\ 3^{-2 n}}= \frac{1}{8}\ \sum_{n=1}^{N} 3^{2n} = \frac{9}{8}\ \frac{1-9^{N}}{1-9} = \frac{9}{64}\ (9^{n}-1)

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by Punch View Post
    Express \sum_{n = 1}^N(\frac{1}{8(3^{-2n})}) in terms of N. but ans is \frac{9}{64}(9^N-1)
    Write \sum_{n = 1}^N(\frac{1}{8(3^{-2n})})=\frac{1}{8}\sum_{n = 1}^N(9^N)

    If S=9+9^2+\cdots+9^N then 9S=9^2+9^3+\cdots+9^{N+1}.

    Subtract 8S=9^{N+1}-9=9(9^N-1})

    Can you finish?
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  5. #5
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    Quote Originally Posted by Plato View Post
    Write \sum_{n = 1}^N(\frac{1}{8(3^{-2n})})=\frac{1}{8}\sum_{n = 1}^N(9^N)

    If S=9+9^2+\cdots+9^N then 9S=9^2+9^3+\cdots+9^{N+1}.

    Subtract 8S=9^{N+1}-9=9(9^N-1})

    Can you finish?
    S=\frac{9(9^N-1)}{8}

    But this answer varies from the answer given in post 3 and my answer sheet. What you said made sense to me but could there be a mistake in the workings?
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  6. #6
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    Quote Originally Posted by Punch View Post
    S=\frac{9(9^N-1)}{8}
    But this answer varies from the answer given in post 3 and my answer sheet. What you said made sense to me but could there be a mistake in the workings?
    You forgot part of it: \frac{1}{8}S=\frac{9(9^N-1)}{64}
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  7. #7
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    thanks!
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