Prove that the generating function 1/(1-z) = (1+z)(1+z^2)(1+z^4)(1+z^8)...
which is also equal to 1+z+z^2+z^3+z^4+... when you multiply out the
binomials.
(1/(1-z))^k = {Sigma [from i=0 to infinity] C(i+k-1, k-1)z^i} also.
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.
Thanks
Message was edited by: Mike
