Having a lot of trouble solving this recurrence with iteration and finding a closed form... I'm learning discrete math and didn't have any trouble with any recurrences in the examples I went over through the chapters on it, but this one problem at the end of the first chapter is killing me, and there's no solution in the book. There's got to be some trick to it that I'm just not seeing. Below is the recurrence and how I'm trying to solve it.

So, right off the bat this one confuses me because $a_0=1$. So, if I try to solve for $a_1$ just by plugging in $a_0$, I get 1. This continues on and and on and on for an as $n$ increases. So, I tried using iteration to go up to $a_5$ to find a pattern. 
I get:
$a_5 = 3(a_4)-2 = 3a_0^5-2(3^4)-2(3^3)-2(3^2)-2(3)-2$. Now, this kind of looks like a geometric sequence, but it's all subtracting vs adding, and if I plug one in for $a_0$, I get $1-2(3^4)-2(3^3)-2(3^2)-2(3)-2$. Is my closed form then $a_n=1-2(3^n-1)$? This just doesn't seem right to me. 
Then if I try to solve by induction to try and check my answer, I get:
Basis: (used 0 since I don't know 1) for $n=0$, $a_n=1-2(3^{n-1}) = 1-2(3^{0-1}) = 1$ which is true.
NTS: $n_K+1=1-2(3^{k+1-1})$
so, $n_k+1=3a_k-1-2
        =3(1-2(3^k-1))-2$  <-- and then I get stuck here, so I can't verify by induction.
Thanks, any input is highly appreciated. I'm ripping my hair out over this.

Using the methods from Wilf's "generatingfunctionology", define the ordinary generating function $A(z) = \sum_{n \ge 0} a_n z^n$. If you write the recurrence as:
$$
a_{n + 1} = 3 a_n - 2
$$
you see that the recurrence is valid for $n \ge 0$. Multiply by $z^n$ and sum for $n \ge 0$:
$$
\sum_{n \ge 0} a_{n + 1} z^n = 3 \sum_{n \ge 0} a_n z^n - 2 \sum_{n \ge 0} z^n
$$
The first sum is $\frac{A(z) - a_0}{z}$, the second is $A(z)$, the third a geometric series:
$$
\frac{A(z) - 1}{z} = 3 A(z) - 2 \frac{1}{1 - z}
$$
This gives:
$$
A(z) = \frac{1}{1 - z}
$$
and so $a_n = 1$.
