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Math Help - Evaluation a non-linear series?

  1. #1
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    Evaluation a non-linear series?

    How do I evaluate \sum_{n=1}^{20} (2^n - 1) without havng to add up each element?

    Thanks
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  2. #2
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    Quote Originally Posted by chancey
    How do I evaluate \sum_{n=1}^{20} (2^n - 1) without havng to add up each element?

    Thanks
    This is a geometric series.
    ---
    (2^1+2^2+...+2^{20})-(1+1...+1)
    Thus,
    2(1+2+2^2+...+2^{19})-20
    Use geometric sum formula,
    2(2^{20}-1)-20
    Simplify,
    2^{21}-22
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  3. #3
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    Quote Originally Posted by ThePerfectHacker
    This is a geometric series.
    ---
    (2^1+2^2+...+2^{20})-(1+1...+1)
    Thus,
    2(1+2+2^2+...+2^{19})-20
    Use geometric sum formula,
    2(2^{20}-1)-20
    Simplify,
    2^{21}-22
    Right, but thats not the answer in the back of the book...

    I did it this way:
    \sum_{n=1}^{20} (2^n - 1) = \sum_{n=1}^{20} (2^n) - 20

    \sum_{n=1}^{20} (2^n) = \frac{1-2^{21}}{1-2} - 1 = 2097150

    \because \sum_{n=0}^{k} (ar^n) = \frac{a(1-r^{k+1})}{1-r}

    \therefore \sum_{n=1}^{20} (2^n) - 20 = 2097150 - 20 = 2097130

    Which is the same answer as yours, however the answer in the book says 2097170. They must of added the 20 instead of subtracted it. OK thanks
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  4. #4
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    Quote Originally Posted by chancey
    Right, but thats not the answer in the back of the book...

    I did it this way:
    \sum_{n=1}^{20} (2^n - 1) = \sum_{n=1}^{20} (2^n) - 20

    \sum_{n=1}^{20} (2^n) = \frac{1-2^{21}}{1-2} - 1 = 2097150

    \because \sum_{n=0}^{k} (ar^n) = \frac{a(1-r^{k+1})}{1-r}

    \therefore \sum_{n=1}^{20} (2^n) - 20 = 2097150 - 20 = 2097130

    Which is the same answer as yours, however the answer in the book says 2097170. They must of added the 20 instead of subtracted it. OK thanks
    Then the book is inncorect.
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