<s>[INST] Without using characteristic equation, how can I show genenal formula of this sequence? I know that, solution of the sequence $x_n$ so that 
$$x_{n+2}-2 x_{n+1}= 2  x_{n+1} - 4 x_{n}$$
is $x_n = (A+ Bn)2^n$. I want to show this without using characteristic equation. I tried
\begin{align*}
   x_{n+2}-2 x_{n+1} &= 2 x_{n+1} - 4 x_{n} \\
   &= 2(x_{n+1}-2x_{n}) \\
   &= 2^2(x_{n}-2x_{n-1}) \\
   &= \ldots \ldots \\
   &= 2^n (x_2 - 2x_1)
\end{align*}
From here, I cannot get the solution. Without using characteristic equation, how can I show general formula of this sequence?
 [/INST] $\begin{align}x_{n+2}=2x_{n+1}+2^n(x_2-2x_1)&&2x_{n+1}=2[2x_n+2^{n-1}(x_2-2x_1)]=4x_{n}+2^n(x_2-2x_1)\\
x_{n+2}=4x_n+2^n(x_2-2x_1)2&&4x_n=4[2x_{n-1}+2^{n-2}(x_2-2x_1)]=8x_{n-1}+2^n(x_2-2x_1)\\
x_{n+2}=8x_{n-1}+2^n(x_2-2x_1)3&&8x_{n-1}=8[2x_{n-2}+2^{n-3}(x_2-2x_1)]=16x_{n-2}+2^n(x_2-x_1)\end{align}$
$\dots$
$x_{n+2}=2^nx_2+2^n(x_2-2x_1)n$
i.e., $x_n=2^{n-2}x_2+2^{n-2}(x_2-2x_1)(n-2)$
</s>