2. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
\frac{a_{n+1}+a_{n}-1}{a_{n+1}-a_{n}+1}=n\left(n \in \mathbf{Z}_{+}\right) \text {, and } a_{2}=6 \text {. }
$$

Then the general term formula of the sequence $\left\{a_{n}\right\}$ is $a_{n}=$ $\qquad$
2. $n(2 n-1)$.

From the given information, we have
$$
\begin{array}{l}
\frac{a_{n}-1}{a_{n+1}}=\frac{n-1}{n+1} \\
\Rightarrow \frac{a_{n+1}}{n+1}=\frac{a_{n}}{n-1}-\frac{1}{n-1} \\
\Rightarrow \frac{a_{n+1}}{(n+1) n}=\frac{a_{n}}{n(n-1)}-\frac{1}{n(n-1)} \\
\Rightarrow \frac{a_{n+1}}{(n+1) n}-\frac{1}{n}=\frac{a_{n}}{n(n-1)}-\frac{1}{n-1} \\
\Rightarrow \frac{a_{n}}{n(n-1)}-\frac{1}{n-1}=\frac{a_{2}}{2 \times 1}-1=2 \\
\Rightarrow a_{n}=n(2 n-1) .
\end{array}
$$