How to solve the recurrence relation $a_n - 2 a_{n-1} = 3 \times 2^n, a_0 = 1$ How to solve the recurrence relation $a_n - 2 a_{n-1} = 3 \times 2^n, a_0 = 1$. By looking at the terms of the relation, it can be seen that it is linear in nature but it is not homogeneous. How to solve such a recurrence relation?

Hint:
Let $a_m=am2^m+b_m$
$$3\cdot2^n=a_n-2a_{n-1}=an2^n+b_n-2\{a(n-1)2^{n-1}+b_{n-1}\}=a2^n+b_n-2_{n-1}$$
Set $a=3$ to find $b_n-2_{n-1}=0$
