# Prove 1/(z-1) by induction

1. Mar 3, 2008

### nowimpsbball

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 = {$$\Sigma$$[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. Mar 3, 2008

### e(ho0n3

So you want to prove that 1/(1-z) = (1+z)(1+z2)(1+z4)...?

3. Mar 4, 2008

### HallsofIvy

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