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.

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:
$$ a_n = \color{red}{3\cdot\frac{2^{n-1}}{3^{n-1}}-1}.\tag{3}$$
