particular solution of a difference equation I am unable to find a particular solution of the following difference equation
$$
y[k-1]-5y[k]+6y[k+1]=-u[k-1]+4u[k]
$$
with $u[k]=\big(\frac{1}{2}\big)^k$.
This is what I tried so far. Because $u[k]=\big(\frac{1}{2}\big)^k$ we try $y_p[k]=\alpha \big(\frac{1}{2}\big)^k$. Substituting this is in the difference equation gives us
$$
\begin{align}
\alpha \bigg(\frac{1}{2}\bigg)^{k-1}-5\alpha \bigg(\frac{1}{2}\bigg)^k + 6\alpha \bigg(\frac{1}{2}\bigg)^{k+1} &= -\bigg(\frac{1}{2}\bigg)^{k-1} + 4 \bigg(\frac{1}{2}\bigg)^k\\
\bigg(\frac{1}{2}\bigg)^{k-1}\bigg(\alpha -\frac{5\alpha}{2}+\frac{6\alpha}{4}\bigg)&= \bigg(\frac{1}{2}\bigg)^{k-1}
\end{align}
$$
Thus $\Big(\alpha -\frac{5\alpha}{2}+\frac{6\alpha}{4}\Big)=1$ which results in $0\alpha=2$ which has no possible solution.
The problem is that I have no idea which particular solution I could try that solves this problem, namely the fact that $1-5/2+6/4=0$.
Any help would be greatly appreciated.

We change notation slightly. Note that $-u_{k-1}+4u_k=\frac{4}{2^k}-\frac{2}{2^k}=\frac{2}{2^k}$. We look for a solution of the shape $y_i=\alpha\frac{i}{2^i}$.
Substitute in $6y_{k+1}-5y_k+y_{k-1}$,  forgetting about the $\alpha$ for a while. We get
$$\frac{6(k+1)}{2^{k+1}}-\frac{5k}{2^k}+\frac{k-1}{2^{k-1}}.$$
This simplifies to 
$$\frac{1}{2^{k+1}}\left(6k+6 -10k+4k+4k-4\right).$$
I am sure you can finish from here.
