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Thread: Evaluation a non-linear series?

  1. #1
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    Evaluation a non-linear series?

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

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

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

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

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

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

    Which is the same answer as yours, however the answer in the book says $\displaystyle 2097170$. 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:
    $\displaystyle \sum_{n=1}^{20} (2^n - 1) = \sum_{n=1}^{20} (2^n) - 20$

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

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

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

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