Use generating functions to solve the recurrence relation Use generating functions to solve $a_n = 3a_{n-1} - 2a_{n-2} + 2^n + (n+1)3^n$.
What I have so far, not sure if I forgot to do something or am missing out on something obvious:
Define $$G(x) = \sum_{n=0}^{\infty} a_nx^n$$ 
Then, $G(x) = a_0 + a_1x + a_2x^2 + \sum_{n=3}^{\infty} a_nx^n$ 
= $a_0 + a_1x + a_2x^2 + \sum_{n=3}^{\infty} (3a_{n-1} - 2a_{n-2} + 2^n + (n+1)3^n)x^n$
= $a_0 + a_1x + a_2x^2 + \sum_{n=3}^{\infty} 3a_{n-1}x^n - \sum_{n=3}^{\infty} 2a_{n-2}x^n + \sum_{n=3}^{\infty} 2^nx^n + \sum_{n=3}^{\infty} (n+1)3^nx^n$
= $a_0 + a_1x + a_2x^2 + 3\sum_{n=3}^{\infty} a_{n-1}x^n - 2\sum_{n=3}^{\infty} a_{n-2}x^n + \sum_{n=3}^{\infty} (2x)^n + \sum_{n=3}^{\infty} (n+1)(3x)^n$
=$a_0 + a_1x + a_2x^2 + 3x\sum_{n=3}^{\infty} a_{n-1}x^{n-1} - 2x^2\sum_{n=3}^{\infty} a_{n-2}x^{n-2} + \sum_{n=3}^{\infty} (2x)^n +  
 \sum_{n=3}^{\infty} (n+1)(3x)^n$
= $a_0 + a_1x + a_2x^2 + 3x\sum_{n=2}^{\infty} a_nx^n - 2x^2\sum_{n=1}^{\infty} a_nx^n + \sum_{n=3}^{\infty} (2x)^n +  
 \sum_{n=3}^{\infty} (n+1)(3x)^n$
 A: After making these replacements, you will have $G(x)$ on both sides, and a couple of other summations you can (hopefully?) find closed forms for. Then you can solve for $G(x)$.
$$
\sum_{n=1}^\infty a_nx^n=G(x)-a_0\hspace{1.3cm}
$$
$$
\sum_{n=2}^\infty a_nx^n=G(x)-a_0-a_1x
$$
A: Hint: if $G(x)$ is the generating function for $a_n$, what are $x G(x)$ and $x^2 G(x)$ the generating functions for?
A: Given that $G(x) = \sum_{n=0}^{\infty} a_nx^n$, you can write $a_n$ as $[x^n]G(x)$. Then the recurrence becomes $$[x^n]G(x) = 3[x^{n-1}]G(x) - 2[x^{n-2}]G(x) + 2^n + (n+1)3^n$$
If we introduce functions $P(x) = \sum_{n=0}^\infty 2^n x^n$ and $Q(x) = \sum_{n=0}^\infty (n+1)3^n x^n$ then, by the linearity of coefficient extraction, $$[x^n]G(x) = 3[x^n]xG(x) - 2[x^n]x^2G(x) + [x^n]P(x) + [x^n]Q(x)$$ and if this holds for every $n$ then $$G(x) = 3xG(x) - 2x^2G(x) + P(x) + Q(x)$$ which rearranges to $$G(x) = \frac{P(x) + Q(x)}{2x^2 - 3x + 1}$$
Finding closed forms for $P$ and $Q$ is left as an exercise...
