Recurrence relation $a_r+6a_{r-1}+9a_{r-2}=3$, then find $a_{20}$ 
Consider the recurrence relation $a_r+6a_{r-1}+9a_{r-2}=3$, given that $a_0=0, a_1=1$. Let $a_{20}=x\times10^9$, then the value of $x$ is______ .


My attempt:
$a_r=3-6a_{r-1}-9a_{r-2}$
I calculated manually, I get $a_{20}=-6465079410=-6.46\times10^9$, so $x=-6.46$

Can you explain in formal way, please? 


Let $f(z)=\sum_{n=0}^\infty a_n z^n$. Multiplying both sides of the recurrence by $z^n$ and summing over $n$ yields
$$\sum_{n=2}^\infty a_n z^n +6\sum_{n=2}^\infty a_{n-1}z^n +9\sum_{n=2}^\infty a_{n-2}z^n = 3\sum_{n=2}^\infty z^n, $$
or
$$f(z)-(a_0+a_1z)+6z(f(z)-a_0) + 9z^2f(z)=\frac{3z^2}{1-z}. $$ Substituting $a_0$ and $a_1$ and isolating $f(z)$, we have $$f(z)(1+6z+9z^2) = \frac{3z^2}{1-z}+z, $$
and hence
$$f(z) = \frac z{(1+3z)^2}+\frac{3z^2}{(1-z)(1+3z)^2}. $$ Partial fraction decomposition yields
$$f(z) = \frac3{16}\left(\frac1{1-z}\right)-\frac5{48}\left(\frac1{1+3z}\right)-\frac1{12}\left(\frac1{(1+3z)^2}\right). $$
Manipulating the geometric series $$\sum_{n=0}^\infty z^n=\frac1{1-z} $$ we find that $$f(z) = \sum_{n=0}^\infty \frac3{16}z^n - \sum_{n=0}^\infty\frac5{48}(-3)^n z^n-\sum_{n=0}^\infty \frac1{12}(n+1)(-3)^nz^n.$$
Simplifying, we have
$$f(z) = \sum_{n=0}^\infty\left(\frac3{16}-\frac5{48}(-1)^n3^{n}-\frac1{12}(n+1)(-1)^n3^{n} \right)z^n, $$
and so we conclude that $$a_n=\frac3{16}-\frac5{48}(-1)^n3^{n}-\frac1{12}(n+1)(-1)^n3^{n}. $$
In particular,
$$a_{20} = \frac{3}{16}-\frac{5}{48} (-1)^{20} 3^{20}-\frac{21}{12} (-1)^{20} 3^{20}=-6465079410. $$
