2. The infinite sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1}=3 a_{n}-4$ $\left(n \in \mathbf{N}_{+}\right)$, and $\left\{a_{n}\right\}$ is a bounded sequence. Then the general term of the sequence $a_{n}=$ $\qquad$ .
2. $a_{n}=2$.

Transform $a_{n+1}=3 a_{n}-4$ into
$$
a_{n+1}-2=3\left(a_{n}-2\right) \text {. }
$$

Assume $a_{1} \neq 2$.
From equation (1), $\left\{a_{n}-2\right\}$ is a geometric sequence with a common ratio of 3.
Therefore, $a_{n}=2+\left(a_{1}-2\right) 3^{n-1}$.
When $n \rightarrow \infty$, $3^{n-1} \rightarrow \infty$, hence $a_{n} \rightarrow \infty$, which contradicts the given condition.
Therefore, $a_{1}=2$.
From (1), we know $a_{n}=2$.