I am having trouble solving this problem.
$$ A(z) = \frac{5+2z}{1-4z^2} $$
I know how to solve
$$ A(z) =\frac{1}{5-6z+z^2} $$
$$ A(z) = \frac{1}{(z-1)*(z-5)}= \frac{A}{z-1}+\frac{B}{z-5}$$ $$A = \frac{-1}{4}\quad B=\frac{1}{4}$$ $$A(z)=\frac{-1}{4*(z-1)}+\frac{1}{4*(z-5)}$$ $$ A(z)=\frac{1}{4*(1-z)}-\frac{1}{20*(1-z/5)}$$ $$ A(z) =\frac{1}{4}*\sum_{r=0}^\infty(z^r)-\frac{1}{20}*\sum_{r=0}^\infty((\frac{z}{5})^r) $$ $$ A(z)= \frac{1}{4} \left[\sum_{r=0}^\infty \left(1-\frac{1}{5^{r+1}}\right){z^r}\right] $$ Since $$A(z)=\sum_{r=0}^\infty({a_r}*{z^r}) $$
therefore $$ a_r= \frac{1}{4}\sum_{r=0}^\infty({1-\frac{1}{5^{r+1}}})$$
I need answer in similar fashion