General term of recurrence relation Find the general term of the following recurrence relation:
$$a_{1} = 2$$
$$a_{n+1}  = \frac{2a_{n} - 1}{3}$$
I've tried to find the first few terms:
$$a_{1} = 2$$
$$a_{2} = \frac{2 \cdot 2 - 1}{3} = 1$$
$$a_{3} = \frac{2 \cdot 1 - 1}{3} = \frac{1}{3}$$
$$a_{4} = -\frac{1}{9}$$
$$a_{5} = -\frac{1}{27}$$
but can't see any pattern, especially considering the first 2 terms.
 A: We have
$$ 3a_{n+1}=2a_n-1 \tag{1}$$
and we may try to get rid of the inhomogeneous term $-1$ by a suitable substitution.
For instance, by setting $a_n=b_n-1$, we get $b_1=3$ and
$$ 3b_{n+1}=2b_n \tag{2} $$
from which $b_n = 3\cdot\frac{2^{n-1}}{3^{n-1}}$ readily follows. The general term of the original sequence is so:
$$ a_n = \color{red}{3\cdot\frac{2^{n-1}}{3^{n-1}}-1}.\tag{3}$$
A: Following your (@Jack D'Aurizio) suggestion, here is what I understood:
The given recurrence relation: $3a_{n} = 2a_{n-1} - 1$, with initial conditions: $a_{1} = 2$ is a $1^{st}$ degree linear non-homogeneous recurrence relation.
Solution: 
We are searching for solution1 of the form: $a_{n} = b_{n} + h_{n}$ (1), where $h_{n}$ is the sequence satisfying the associated homogeneous recurrence relation and $b_{n}$ is a solution which is similar to $f(n)$.
1.The associated homogeneous recurrence relation is: $3h_{n} = 2h_{n-1}$ and its characteristic equation is: $3r - 2 = 0$ or $ r = \frac{2}{3}$ and the solution is:
$$h_{n} = \alpha_{0} (\frac{2}{3})^{n-1} $$  
From the initial conditions: $a_{1} = \alpha_{0} (\frac{2}{3})^{1} = 2$ or $\alpha_{0} = 3$, so the solution of the homogeneous recurrence is:
$$h_{n} = 3 (\frac{2}{3})^{n-1} $$  


*In our case $b_{n} = -1$ and from (1) we have:


$$a_{n} = 3 (\frac{2}{3})^{n-1}  - 1$$

1. $a_{n}$ satisfies both the recurrence relation and the initial conditions.
2. $f(n)$ - function depending only on $n$.
