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Math Help - sigma notation

  1. #1
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    sigma notation

    Hi.

    I have this example in my boo

    I have a similar question which I started,
    But I don't really understand how the sigma notation worked in the first link... therefore I don't know the next line on mine.

    Looking for any pointers/help to this next line

    Many thanks

    tom
    Last edited by Tom123; November 7th 2007 at 06:36 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Tom123 View Post
    Hi.

    I have this example in my book http://img229.imageshack.us/img229/686/math1tm3.jpg

    I have a similar question which I started, ImageShack - Hosting :: math2um1.jpg

    But I don't really understand how the sigma notation worked in the first link... therefore I don't know the next line on mine.

    Looking for any pointers/help to this next line

    Many thanks

    tom
    Let's take the first example step by step.
    \sum_{i = 1}^n2^{n - i}(i)2^i

    = \sum_{i = 1}^n2^{n - i}2^i(i)

    = \sum_{i = 1}^n2^{n - i + i}(i)

    = \sum_{i = 1}^n2^n(i)

    = 2^n \sum_{i = 1}^n(i) <-- The 2^n does not depend on i so it is constant with respect to the summation

    = 2^n \cdot \frac{n(n + 1)}{2}

    -Dan
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  3. #3
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    Alright, so the step is simply this?


    4^n n(n+1)
    ........4
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Tom123 View Post
    Alright, so the step is simply this?


    4^n n(n+1)
    ........4
    You mean in your problem? Not quite, because
    \sum_{i = 1}^n 4^{n - i}(i) 4^i = 4^n \sum_{i = 1}^ni = 4^n \cdot \frac{n(n + 1)}{2}

    -Dan
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