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Math Help - Fill in missing step please!

  1. #1
    Junior Member
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    Fill in missing step please!

    Hi, I am probably an idiot for missing this but im sick of staring at this problem i thought i would come here to be enlightened. anyways
    phase 1
    assuming n,m > 0 and real numbers and m > n
    \frac{1}{2^n}\left(1+\frac{1}{2}+...+\frac{1}{2^{m-n-1}}\right)
    phase 2
    \frac{1}{2^n}\cdot\frac{\left(1-(\frac{1}{2})^{m-n-1+1}\right)}{1-\frac{1}{2}}

    there are all equal statements im told, but I have no clue the 'trick' to widdle this down from phase 1 to phase 2. thanks in advance
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  2. #2
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    Let S = 1 + r + r^2 + \cdots + r^n. Then  1+rS = 1 + r + r^2 + \cdots + r^n + r^{n+1} = S + r^{n+1} and rearranging gives S = {1-r^{n+1}\over1-r}. Use this (with r=1/2) to get from phase 1 to phase 2.
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  3. #3
    Super Member Failure's Avatar
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    Quote Originally Posted by fizzle45 View Post
    Hi, I am probably an idiot for missing this but im sick of staring at this problem i thought i would come here to be enlightened. anyways
    phase 1
    assuming n,m > 0 and real numbers and m > n
    \frac{1}{2^n}\left(1+\frac{1}{2}+...+\frac{1}{2^{m-n-1}}\right)
    phase 2
    \frac{1}{2^n}\cdot\frac{\left(1-(\frac{1}{2})^{m-n-1+1}\right)}{1-\frac{1}{2}}

    there are all equal statements im told, but I have no clue the 'trick' to widdle this down from phase 1 to phase 2. thanks in advance
    If n,m are, in addition, integers, then this is just an application of the formula for the sum of a geometric series:
    1+q+q^2+\cdots+q^N=\frac{1-q^N}{1-q}
    were q=\frac{1}{2} and N=m-n-1
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  4. #4
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    thank for quick response guys very helpful!
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