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Math Help - Geometric Progression and Summation

  1. #1
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    Geometric Progression

    The sum of the first n terms of a series is given by the expression (1-3^{-2n}).

    Find the least value of k such that the sum of the terms from the k^{th} term is less than \frac{1}{3000}

    1-3^{2k} < \frac{1}{3000}

    -3^{-2k} < -\frac{2999}{3000}

    3^{-2k} > \frac{2999}{3000}

    -2k < \frac{lg\frac{2999}{3000}}{lg3}

    k > 0.0001517

    but ans is 5
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  2. #2
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    It doesn't say to sum the terms up to the \displaystyle k^{\textrm{th}} term, it says FROM the \displaystyle k^{\textrm{th}} term, presumably up to the \displaystyle n^{\textrm{th}} term.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    It doesn't say to sum the terms up to the \displaystyle k^{\textrm{th}} term, it says FROM the \displaystyle k^{\textrm{th}} term, presumably up to the \displaystyle n^{\textrm{th}} term.
    thanks for pointing that out! i'll attempt the question again with this in mind.
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  4. #4
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    \sum_{n = k}^N(1-3^{-2n}) < \frac{1}{3000}

    \sum_{n = k}^N1-\sum_{n = k}^N(3^{-2n}) < \frac{1}{3000}

    (N-k+1)(1)-(3^{-2k}+3^{-2(k+1)}+...+3^{-2N}) < \frac{1}{3000}

    N-k+1-\frac{3^{-2k}(1-9^{N-k+1})}{1-9}<\frac{1}{3000}

    but N doesnt seem to be cancelling out
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  5. #5
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    Quote Originally Posted by Punch View Post
    \sum_{n = k}^N(1-3^{-2n}) < \frac{1}{3000}

    \sum_{n = k}^N1-\sum_{n = k}^N(3^{-2n}) < \frac{1}{3000}

    (N-k+1)(1)-(3^{-2k}+3^{-2(k+1)}+...+3^{-2N}) < \frac{1}{3000}

    N-k+1-\frac{3^{-2k}(1-9^{N-k+1})}{1-9}<\frac{1}{3000}

    but N doesnt seem to be cancelling out
    am i on the right track?
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