Solve the non-homogenous recurrence $h_n = 3h_{n-1} + 3(2)^n, h_0 = 1$? I am trying to solve the non-homogenous recurrence $h_n = 3h_{n-1} + 3(2)^n, h_0 = 1$.
I can see that the solution to the homogenous $h_n = 3h_{n-1}$ is clearly $h_n = 3^n$. That being said, I have no idea how to solve $h_n - 3h_{n-1} = 3(2)^n$: I assume it's some solution of the sort $(c_1n + c_2)(2)^n$, but I don't know where we'd go from here! Can anyone give advice?

Let $ n\in\mathbb{N}^{*} $, note that for all $ k\in\mathbb{N} $, we have : \begin{aligned}h_{n-k}&=3h_{n-k-1}+3\times 2^{n-k}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3^{k}h_{n-k}&=3^{k+1}h_{n-k-1}+3^{k+1}\times 2^{n-k}\\ \Longrightarrow\sum_{k=0}^{n-1}{\left(3^{k}h_{n-k}-3^{k+1}h_{n-k-1}\right)}&=3\times 2^{n}\sum_{k=0}^{n-1}{\left(\frac{3}{2}\right)^{k}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h_{n}-3^{n}h_{0}&=3\times 2^{n}\times\frac{\left(\frac{3}{2}\right)^{n}-1}{\frac{3}{2}-1}\\ \iff\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  h_{n}&=3^{n}+6\left(3^{n}-2^{n}\right)\end{aligned}
