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Answer:

The answer is "[tex]x_k= -\frac{9}{16} (-2)^k + \frac{9}{16} 2^k +\frac{3}{8} k\times 2^k\\\\[/tex]"

Step-by-step explanation:

[tex]\to F(Z)=\frac{3z(z-1)}{z^3-2z^2-4z+8}\\\\\to \frac{F(Z)}{z}=\frac{3z(z-1)}{z(z^3-2z^2-4z+8)}\\\\\to \frac{F(Z)}{z}=\frac{3(z-1)}{(z^3-2z^2-4z+8)}\\\\\to \frac{F(Z)}{z}=-\frac{9}{16} \frac{1}{z+2} + \frac{9}{16} \frac{1}{z-2} +\frac{3}{4} \frac{1}{(z-2)^2}\\\\\to F(Z)=-\frac{9}{16} \frac{z}{z+2} + \frac{9}{16} \frac{z}{z-2} +\frac{3}{4} \frac{z}{(z-2)^2}\\\\\to x_k= -\frac{9}{16} (-2)^k + \frac{9}{16} 2^k +\frac{3}{8} k\times 2^k\\\\[/tex]

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