I've split it up into partial fractions and got $1/z$ - $2/(z-1)$ + $1/(z-2)$ but I'm unsure sure what to do now.
I think I have done part $(i)$. I get
$$z^{-1} + \sum_{n=0}^\infty (2-(1/2^{n+1})z^n)$$
What do I do with the $z^{-1}$? How do I get it into the summation?
For part $(ii)$ I get ($\sum_{n=0}^\infty (-1)^n (z-1)^n) - 2/(z-1) - (\sum_{n=0}^\infty (z-1)^n)$ Is this right? How do I make it look nicer?
For part $(iii)$ I am confused. I have to make the denominator so it is of the form $1-(1/z)$? What do I do with the $1/z$ part of the partial fractions?
So far, for part $(iv)$ I have ($\sum_{n=0}^\infty (-1)^n (2)^{n} /(z-2)^{n+1})$ - 2($\sum_{n=0}^\infty (-1)^n /(z-2)^{n+1}$) + $1/(z-2)$
Thanks