China Mathematical Competition

(1) The given expression can be rewritten as
$$a_{n+1} = \frac{2(t^{n+1} - 1)(a_n + 1)}{a_n + 2t^n - 1} - 1.$$
Then
$$\frac{a_{n+1} + 1}{t^{n+1} - 1} = \frac{2(a_n + 1)}{a_n + 2t^n - 1} = \frac{\frac{2(a_n + 1)}{t^n - 1}}{\frac{a_n + 1}{t^n - 1} + 2}.$$
Let $\frac{a_n + 1}{t^n - 1} = b_n$. Then $b_{n+1} = \frac{2b_n}{b_n + 2}$, with $b_1 = \frac{a_1 + 1}{t - 1} = \frac{2t - 2}{t - 1} = 2$.

Furthermore, $\frac{1}{b_{n+1}} = \frac{1}{b_n} + \frac{1}{2}$, $\frac{1}{b_1} = \frac{1}{2}$. Then
$$\frac{1}{b_n} = \frac{1}{b_1} + (n-1) \cdot \frac{1}{2} = \frac{n}{2}.$$
Therefore, $\frac{a_n + 1}{t^n - 1} = \frac{2}{n}$, which means $a_n = \frac{2(t^n - 1)}{n} - 1$.

(2) We have
$$\begin{aligned} a_{n+1} - a_n &= \frac{2(t^{n+1} - 1)}{n+1} - \frac{2(t^n - 1)}{n} \\ &= \frac{2(t-1)}{n(n+1)} \left[ n(1+t+\cdots+t^{n-1}+t^n) - (n+1)(1+t+\cdots+t^{n-1}) \right] \\ &= \frac{2(t-1)}{n(n+1)} \left[ nt^n - (1+t+\cdots+t^{n-1}) \right] \\ &= \frac{2(t-1)}{n(n+1)} \left[ (t^n-1) + (t^n-t) + \cdots + (t^n-t^{n-1}) \right] \\ &= \frac{2(t-1)^2}{n(n+1)} \left[ (t^{n-1}+t^{n-2}+\cdots+1) + t(t^{n-2}+t^{n-3}+\cdots+1) + \cdots + t^{n-1} \right]. \end{aligned}$$
It is obvious that $a_{n+1} - a_n > 0$ for $t > 0$ ($t \neq 1$). Therefore, $a_{n+1} > a_n$.