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Prove 1/(z-1) by induction

  1. Mar 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that the generating function 1/(1-z) = (1+z)(1+z^2)(1+z^4)(1+z^8)...
    which is also to 1+z+z^2+z^3+z^4+... when you multiply out the binomials.


    2. Relevant equations
    (1/(1-z))^k = {[tex]\Sigma[/tex][from i=0 to infinity] C(i+k-1, k-1)z^i}


    3. The attempt at a solution
    I've been playing around with this for a while. I thought I had it, but then I realized that I plugged the "n+1" into the exponent...ie 1+z+z^2+z^3+z^4+...+z^n+z^n+1....but that is not right...I need to plug in z+1 into each z.

    PS Ignore the topic name, it should be 1/(1-z)

    Thanks
    Thanks
     
    Last edited: Mar 3, 2008
  2. jcsd
  3. Mar 3, 2008 #2
    So you want to prove that 1/(1-z) = (1+z)(1+z2)(1+z4)...?
     
  4. Mar 4, 2008 #3

    HallsofIvy

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    Why do you refer to induction? There is no "n" in your formula to do an induction on!
     
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