Proof by Induction

1. Sep 4, 2011

83956

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. Sep 4, 2011

lanedance

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)

3. Sep 4, 2011

83956

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
4. Sep 4, 2011

Dick

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.