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Thread: Summation

  1. #1
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    Summation

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

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

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

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

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

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

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

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

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

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

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

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

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

    Kind regards

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

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

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

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

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

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

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

    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
    $\displaystyle S=\frac{9(9^N-1)}{8}$
    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: $\displaystyle \frac{1}{8}S=\frac{9(9^N-1)}{64}$
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  7. #7
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    thanks!
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