Can someone tell me how to find the Z Tranform of the sequence:
$x(n)=n, \quad n=0,1,2,3,4,5 \Rightarrow X(z)=\sum\limits_{n=0}^{N-1}nz^{-n} $
I have searched everywhere I could, but in every single example there is a coefficient raised in $n$ which made Z Transform too easy. I have found the result for something similar to this one in a lot of places but not the proof!
Let's try to manipulate that sum as follows:
[1] $X(z) = z^{-1}+2z^{-2}+3z^{-3}+\cdots+(N-1)z^{-(N-1)}$
[2] $z^{-1}X(z) = z^{-2}+2z^{-3}+3z^{-4}+\cdots+(N-1)z^{-N}$ (Multiply [1] by $z^{-1}$)
[3] $(1-z^{-1})X(z) = z^{-1}+z^{-2}+z^{-3}+\cdots+z^{-(N-1)} - (N-1)z^{-N}$ (Subtract [2] from [1])
[4] $z^{-1}(1-z^{-1})X(z) = z^{-2}+z^{-3}+z^{-4}+\cdots+z^{-N} - (N-1)z^{-(N+1)}$ (Multiply [3] by $z^{-1}$)
[5] $(1-z^{-1})^2X(z) = z^{-1}-Nz^{-N}+(N-1)z^{-(N+1)}$ (Subtract [4] from [3])
[6] $X(z) = \dfrac{z^{-1}-Nz^{-N}+(N-1)z^{-(N+1)}}{(1-z^{-1})^2}$ (Divide both sides of [5] by $(1-z^{-1})^2$)
You should double check the algebra.