Find the inverse transform
\begin{equation} X(z)=\frac{z^3+2z^2-2z}{(z-2)(z^2+2)} \end{equation}
We solve the partial fractions for:
\begin{equation} \frac{X(z)}{z}=\frac{z^2+2z-2}{(z-2)(z^2+2)}=z\bigg(\frac{1}{z-2}+\frac{2}{2+z^2}\bigg) \end{equation}
We now have to inverse transform:
\begin{equation} \frac{z}{z-2}+\frac{2z}{2+z^2} \end{equation}
From the table we have:
\begin{equation} \begin{array} -\frac{z}{z-2} \rightarrow 2^{n}\\ \frac{2z}{z^2+2} \rightarrow ? \end{array} \end{equation}
For the last transform, I did the partial fractions decomposition on it and got
\begin{equation*} \frac{2z}{z^2+2} = \frac{\big(1+\frac{\sqrt{2}i}{4}\big)}{z-\sqrt{2}i}+\frac{\big(1-\frac{\sqrt{2}i}{4}\big)}{z+\sqrt{2}i} \end{equation*}
Using the Residue theorem:
\begin{equation} Res(-\sqrt{2}i)=\lim_{z\rightarrow -\sqrt{2}i}(z-\sqrt{2}i)\frac{\big(1+\frac{\sqrt{2}i}{4}\big)}{z-\sqrt{2}i}z^{n-1}=(1+\frac{\sqrt{2}}{4}i)(-\sqrt{2}i)^{n-1} \end{equation}
and
\begin{equation} Res(\sqrt{2}i)=\lim_{z\rightarrow \sqrt{2}i}(z+\sqrt{2}i)\frac{\big(1-\frac{\sqrt{2}i}{4}\big)}{z+\sqrt{2}i}z^{n-1}=(1-\frac{\sqrt{2}}{4}i)(\sqrt{2}i)^{n-1} \end{equation}
So
\begin{equation} V(n)=(1+\frac{\sqrt{2}}{4}i)(-\sqrt{2}i)^{n-1}+(1-\frac{\sqrt{2}}{4}i)(\sqrt{2}i)^{n-1}-2^n \end{equation}
But Wolframalpha shows something quite different: link which appears as this:
while my solution appears as this:
