Solution of the recurrence $a_n-3a_{n-1}=2^n$, $a_0=5$

Question

What is the solution of the recurrence $a_n-3a_{n-1}=2^n$, $a_0=5$.

I think the solution is $$a_n=3^n(5)+\sum_{i=1}^{n-1}3^{n-i}2^i+2^n.$$ Am I right? Can it be simplified further? Thanks beforehand.

Answer 1

Yes, you are right, your formula can be further simplified. Note that the sum $\sum_{i=1}^{n-1}3^{n-i}2^i=3^n\sum_{i=1}^{n-1}(2/3)^i$ is geometric. Finally you should obtain $$a_n=A\cdot 3^n+B\cdot 2^n$$ where $A$ and $B$ are constants to be determined. Can you take it from here?