# 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!

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook