Solve $a_n=7a_{n-1}-12a_{n-2}+3^n$

Question

$a_0=1\:a_1=2$

Using generating functions I get:

$f\left(x\right)-1-2x=7xf\left(x\right)-7x-12x^2f\left(x\right)+\frac{1}{1+3x}-1-3x$

$f\left(x\right)=\frac{3}{14\left(1+3x\right)}-\frac{13}{2\left(3x-1\right)}\:+\:\frac{40}{7\left(4x-1\right)}$

Now how can I express $a_n$ from this?

Answer 1

Hint. One may just recall that $$ \frac{1}{1-u}=\sum_{n=0}^\infty u^n,\qquad |u|<1, $$ then applying it to $$ f\left(x\right)=\frac{3}{14\left(1+3x\right)}+\frac{13}{2\left(1-3x\right)}\:-\:\frac{40}{7\left(1-4x\right)} $$ respectively with $u$ equals to $-3x$, $3x$ and $4x$.

Can you finish it?

Answer 2

Hint (without generating functions): eliminate $3^n$ between the recurrence relations for $n$ and $n+1$, then you get a $3^{rd}$ order linear homogeneous recurrence:

$$\require{cancel} a_{n+1}-3 a_n=7 a_n-12a_{n-1}+\cancel{3^{n+1}} - 3(7 a_{n-1}-12 a_{n-2}+ \cancel{3^n}) \\[3px] \iff a_{n+1} = 10 a_n - 33 a_{n-1}+36 a_{n-2} $$

The characteristic polynomial of the latter is $t^3-10t^2+33t-36=(t-3)^2(t-4)$ so the solution of the recurrence will be of the form $a_n=(a+b\cdot n)\cdot 3^n+c \cdot 4^n$ where constants $a,b,c$ remain to be determined from the initial conditions.