Proof of 1/(1-x) = 1 + x + x2 + x3 + ... + xn/(1-x) by Induction

[83956]

Homework Statement

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

Homework Equations

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.

 

[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)

 

[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)

 

[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.