Hello all. Can somebody please help with the intermediate steps here? I have reached the point given, but not sure how to advance to the final solution. Many thanks
$\displaystyle \sum_{i=0}^9 (1/2z)^{n}=\frac{1-(2z)^{-10}}{1-(2z)^{-1}}$
Hello all. Can somebody please help with the intermediate steps here? I have reached the point given, but not sure how to advance to the final solution. Many thanks
$\displaystyle \sum_{i=0}^9 (1/2z)^{n}=\frac{1-(2z)^{-10}}{1-(2z)^{-1}}$