Let $x(n)=(n-1)(\frac{1}{2})^{n-2}u(n-2)$, where $u(n-2)$ is shifted unit step function. How can I calculate z transform of this function?
By definition, $X(z)=\sum_{n=-\infty}^{n=\infty}x(n)z^{-n}=\sum_{n=2}^{n=\infty}(n-1)(\frac{1}{2})^{n-2}z^{-n}$.
How to calculate this sum? If there was no $(n-1)$, this would be geometric series and its sum can be easily calculated.
We can write the sum as:
$$\frac1{z} \sum_{n=1}^{\infty} n\left(\frac1{2z}\right)^{n - 1}$$
Now we use (for $|\xi| < 1$):
$$\frac1{(1-\xi)^2} = \sum_{n=1}^{\infty} n\xi^{n-1}$$