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Proof by Induction

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove that 1/(1-x) = 1 + x + x2 + x3 + ... + xn/(1-x) for n>=2

    2. Relevant equations



    3. The attempt at a solution

    I'm not really all that sure how to begin. The base case would be 1/(1-x) = x2/(1-x) and the induction hypothesis would be 1/(1-x) = 1 + x + x2 + x3 + ... + xn/(1-x) but I don't know what the n+1 case is and how to prove that it holds. I guess the n+1 case would logically be 1/(1-x) = 1 + x + x2 + x3 + ... + xn/(1-x) + xn+1/(x-1), but I don't know how to show algebraically that the left hand side equals the right hand side.
     
  2. jcsd
  3. Sep 4, 2011 #2

    lanedance

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    shouldn't the base case be
    1/(1-x) =1+x+ x2/(1-x)

    it might be worth rearranging
    1/(1-x)- x2/(1-x)=1+x

    now try multiplying through by (1-x) on both sides

    now basically you want to show the "n" case implies the "n+1" case to get the induction. To do this either manipulate the n case to show "n+1" case or vice versa
    1/(1-x) = 1 + x + x2 + x3 + ... + xn/(1-x) + xn+1/(x-1)
     
  4. Sep 4, 2011 #3
    Yes, I see the correction in my error of the base case...got it now.

    However, I'm still not getting the n+1 case. So the induction hypothesis is 1/(1-x) = 1 + x + x^2 + ... x^n/(1-x). I want to show 1/(1-x) = 1 + x + x^2 + ... + x^n/(1-x) + x^(n+1)/(1-x). ....

    Basically for the n case 1-x^n = (1-x)(1 + x + x^2 + ... + x^n-1)
     
    Last edited: Sep 4, 2011
  5. Sep 4, 2011 #4

    Dick

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    The n+1 case is 1/(1-x)=1+x+x^2+...+x^n+x^(n+1)/(1-x), isn't it? Take the difference between the "n" case and the "n+1" case and show it's zero.
     
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