Solve the difference equation
\begin{equation} x_{n+1}-2x_n=n \end{equation}
Solution:
I use the z-transform table:
\begin{equation} \begin{array}{cc} First \ term, x_{n+1}:\\ Table\ entry: x_{n+k}\rightarrow zX(z)-\sum_0^{k-1}x_jz^{1-j}= zX(z) \\ Second\ term -2x_n: \\ Table\ entry: cx_n \rightarrow cX(z) =-2X(z)\\ Third\ term, n:\\ Table\ entry: n \rightarrow \frac{z}{(z-1)^2} \end{array} \end{equation}
Therefore the result is
\begin{equation} zX(z)-2X(z)=\frac{z}{(z-1)^2} \end{equation} \begin{equation} X(z)=\frac{z}{(z-1)^2(z-2)} \end{equation}
Is this correct to the knowledge of keen readers?
UPDATE:
Pauls point is given below:
from inverse z-transform table acting on:
\begin{equation} X(z)=\frac{z}{(z-1)^2(z-2)} = \frac{1-2z}{(z-1)^2}+\frac{2}{z-2} \end{equation}
we have:
\begin{equation} \begin{array} f \frac{1-2z}{(z-1)^2}\rightarrow \frac{1}{(z-1)^2}-\frac{2z}{(z-1)^2}\\ \rightarrow z^{-1}\frac{z}{(z-1)^2}-2 \frac{z}{(z-1)^2} \rightarrow (n-1)\sigma_{n-1}-2n, \\ \frac{1}{z-2}\rightarrow 2z^{-1}\frac{z}{z-2}\rightarrow 2^{n} \end{array} \end{equation}
\begin{equation} x_n=(n-1)\sigma_{n-1}-2n+2^{n} \end{equation}
Thanks
You can avoid delayed sequences in your solution by putting
$\frac{X(z)}{z}=\frac{1}{(z-1)^2(z-2)} =\frac{-1}{z-1} -\frac{1}{(z-1)^2}+\frac{1}{z-2}$
Then multiplying through by $z$
$X(z)=\frac{-z}{z-1} -\frac{z}{(z-1)^2}+\frac{z}{z-2}$
which are all in your table