Solve the recurrence $a_n=a_{n-1} + 6a_{n-2} + 5\cdot3^n\; \forall n\geq 2\;\;a_0=4\;\; a_1=6$

Question

Solve the recurrence $a_n=a_{n-1} + 6a_{n-2} + 5\cdot3^n\; \forall n\geq 2\;\;a_0=4\;\; a_1=6$

I let $$g(x)=\sum_{i\geq 0} a_i x^i$$

Using this I was able to obtain

$$g(x)=\frac{39x^2-10x+4}{(3x-1)^2(2x+1)}$$

Now I want the coefficient of $x^n$ in my $g(x)$ which will equal $a^n$. So how do I get this? Also if possible, please verify my $g(x)$ is correct or not. I was searching for online tool to do same, but can't find one

Answer 1

Doing partial fraction decomposition on $g(x)$, we obtain $\frac{3}{1 + 2 x} + \frac{3}{(-1 + 3 x)^2} + \frac{2}{(-1 + 3 x)}$.

Applying Taylor series expansion on the denominator, we obtain

\begin{align}\frac{3}{1 + 2 x} + \frac{2}{(-1 + 3 x)}+ \frac{3}{(-1 + 3 x)^2} &=\sum_{n=0}^\infty (3\cdot(-2)^n+(-2)3^n+)x^n+3^n(3n+3))x^n\\\\ &= \sum_{n=0}^\infty 3^n(3n+1)+3\cdot(-2)^nx^n \end{align}

Therefore, you get $a_n= 3^n(3n+1)+3\cdot(-2)^n$