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 $$\begin{array}{rl}{a}_{n+1}-a_n& =\frac{2(t^{n+1}-1)}{n+1}-\frac{2(t^n-1)}{n}\\ & =\frac{2(t-1)}{n(n+1)}\left[n(1+t+\cdots +t^{n-1}+t^n)-(n+1)(1+t+\cdots +t^{n-1})\right]\\ & =\frac{2(t-1)}{n(n+1)}\left[n{t}^n-(1+t+\cdots +t^{n-1})\right]\\ & =\frac{2(t-1)}{n(n+1)}\left[(t^n-1)+(t^n-t)+\cdots +(t^n-t^{n-1})\right]\\ & =\frac{2(t-1{)}^2}{n(n+1)}\left[(t^{n-1}+t^{n-2}+\cdots +1)+t(t^{n-2}+t^{n-3}+\cdots +1)+\cdots +t^{n-1}\right].\end{array}$$ It is obvious that $a_{n+1}-a_n>0$ for $t>0$ ($t\ne 1$). Therefore, $a_{n+1}>a_n$.