Q5Advanced Engineering Mathematics I

Let $F(z)=\frac{5z}{(2-z)(3z-1)}=\frac{-5z}{(z-2)3(z-1/3)}=\frac{-5/3z}{(z-2)(z-1/3)}$

Divide by z and resolve into partial fractions: $\frac{F(z)}{z}=\frac{-5/3}{(z-2)(z-1/3)}=\frac{A}{z-2}+\frac{B}{z-1/3}$

$A={\frac{-5/3}{z-1/3}|}_{z=2}=\frac{-5/3}{5/3}=-1$ $B={\frac{-5/3}{z-2}|}_{z=1/3}=\frac{-5/3}{-5/3}=1$

So, $\frac{F(z)}{z}=\frac{-1}{z-2}+\frac{1}{z-1/3}⟹F(z)=-\frac{z}{z-2}+\frac{z}{z-1/3}$

Taking the inverse Z-transform using standard pairs ($Z\{a^n\}=\frac{z}{z-a}$): $f(n)=-(2{)}^n+(1/3{)}^n$