Solve the recurrence $a_n - 3a_{n-1} + 2a_{n-2} = 6 \cdot 2^n$
Your calculation is an example of proof by contradiction. You assumed that $a_n = Cn2^n$, and reached a contradiction $0 = 22$. You can conclude that $a_n \neq Cn2^n$.



Your calculation shows that the general solution to your equation is
$$
a_n = A + 2^n B + n2^n C.
$$
For some reason you forgot about the first two terms.



Let $P(x) = \sum_{n=0}^\infty a_n x^n$. Setting $a_{-1} = a_{-2} = 0$, we have
$$
(2x^2-3x+1) P(x) = \sum_{n=0}^\infty x^n (2a_{n-2} - 3 a_{n-1} + a_n) = \\
a_0 + (a_1 - 3a_0) x + \sum_{n=2}^\infty x^n (6 \cdot 2^n) = \\
a_0 + (a_1 - 3a_0) x + 24x^2 \sum_{m=0}^\infty (2x)^m = \\
a_0 + (a_1 - 3a_0) x + \frac{24x^2}{1-2x}.
$$
Since $2x^2-3x+1 = (2x-1)(x-1)$, it follows that
$$
P(x) = \frac{(1-2x)(a_0 + (a_1 - 3a_0) x) + 24x^2}{(1-2x)^2(1-x)} =
\frac{a_0 + (a_1 - 5a_0)x + (24-2a_1+6a_0)x^2}{(1-2x)^2(1-x)}.
$$
Routine calculation shows that
$$
\frac{1}{(1-2x)^2(1-x)} = \sum_{n=0}^\infty (1+2^{n+1}n)x^n,
$$
and so for $n \geq 2$ we get the formula
$$
a_n = a_0 (1 + 2^{n+1}n) + (a_1 - 5a_0) (1 + 2^n(n-1)) + (24 - 2a_1 + 6a_0) (1 + 2^{n-1}(n-2)).
$$
Opening this up, we get the solution
$$
a_n = 24 - a_1 + 2a_0 - (24 - a_1 + a_0) \cdot 2^n + 12 \cdot n2^n.
$$