10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
\begin{array}{l}
a_{1}=2 t-3(t \in \mathbf{R}, \text { and } t \neq \pm 1), \\
a_{n+1}=\frac{\left(2 t^{n+1}-3\right) a_{n}+2(t-1) t^{n}-1}{a_{n}+2 t^{n}-1}\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If $t>0$, compare the size of $a_{n+1}$ and $a_{n}$.
10. (1) From the original equation, we have
$$
\begin{array}{l}
a_{n+1}=-\frac{2\left(t^{n+1}-1\right)\left(a_{n}+1\right)}{a_{n}+2 t^{n}-1}-1 . \\
\text { Then } \frac{a_{n+1}+1}{t^{n+1}-1}=\frac{2\left(a_{n}+1\right)}{a_{n}+2 t^{n}-1}=\frac{\frac{2\left(a_{n}+1\right)}{t^{n}-1}}{\frac{a_{n}+1}{t^{n}-1}+2} .
\end{array}
$$

Let $\frac{a_{n}+1}{t^{n}-1}=b_{n}$. Then
$$
b_{n+1}=\frac{2 b_{n}}{b_{n}+2}, b_{1}=\frac{a_{1}+1}{t-1}=\frac{2 t-2}{t-1}=2 \text {. }
$$

Also, $\frac{1}{b_{n+1}}=\frac{1}{b_{n}}+\frac{1}{2}, \frac{1}{b_{1}}=\frac{1}{2}$, thus,
$$
\begin{array}{l}
\frac{1}{b_{n}}=\frac{1}{b_{1}}+\frac{1}{2}(n-1)=\frac{n}{2} . \\
\text { Hence } \frac{a_{n}+1}{t^{n}-1}=\frac{2}{n} \Rightarrow a_{n}=\frac{2\left(t^{n}-1\right)}{n}-1 .
\end{array}
$$
(2) Notice that
$$
\begin{array}{l}
a_{n+1}-a_{n}=\frac{2\left(t^{n+1}-1\right)}{n+1}-\frac{2\left(t^{n}-1\right)}{n} \\
=\frac{2\left[n t^{n}(t-1)-\left(t^{n}-1\right)\right]}{n(n+1)} \\
=\frac{2(t-1)}{n(n+1)}\left[n t^{n}-\left(1+t+\cdots+t^{n-1}\right)\right] .
\end{array}
$$

Clearly, when $t>0(t \neq 1)$, $t-1$ and $n t^{n}-$ $\left(1+t+\cdots+t^{n-1}\right)$ have the same sign.
Therefore, $a_{n+1}>a_{n}$.