已知数列 $\{a_n\}$ 满足：$a_1=2t-3$（$t\in \mathbb R$ 且 $t\ne \pm 1$），且对任意 $n\in\mathbb N^{\ast}$，有\[a_{n+1}=\dfrac{(2t^{n+1}-3)a_n+2(t-1)t^n-1}{a_n+2t^n-1}.\]
1、求数列 $\{a_n\}$ 的通项公式．
2、若 $t>0$，试比较 $a_{n+1}$ 与 $a_n$ 的大小．
解析
1、由原式变形得$$a_{n+1}=\dfrac{2(t^{n+1}-1)(a_n+1)}{a_n+2t^n-1}-1,$$则$$\dfrac{a_{n+1}+1}{t^{n+1}-1}=\dfrac{2(a_n+1)}{a_n+2t^n-1}=\dfrac{\dfrac{2(a_n+1)}{t^n-1}}{\dfrac{a_n+1}{t^n-1}+2}.$$ 记 $\dfrac{a_n+1}{t^n-1}=b_n$，则 $b_{n+1}=\dfrac{2b_n}{b_n+2}$，且$$b_1=\dfrac{a_1+1}{t-1}=\dfrac{2t-2}{t-1}=2.$$ 又$$\dfrac 1{b_{n+1}}=\dfrac 1{b_n}+\dfrac 12,$$从而有$$\dfrac 1{b_n}=\dfrac 1{b_1}+(n-1)\cdot \dfrac 12 =\dfrac n2,$$故$$\dfrac{a_n+1}{t^n-1}=\dfrac 2n,$$于是有 $a_n=\dfrac{2(t^n-1)}{n}-1$．
2、作差比较得\[\begin{split}a_{n+1}-a_n&=\dfrac{2(t^{n+1}-1)}{n+1}-\dfrac{2(t^n-1)}{n}\\&=\dfrac{2(t-1)}{n(n+1)}\left(n(1+t+\cdots +t^{n-1}+t^n)-(n+1)(1+t+\cdots +t^{n-1})\right)\\&=\dfrac{2(t-1)}{n(n+1)}\left(nt^n-(1+t+\cdots +t^{n-1})\right)\\&=\dfrac{2(t-1)}{n(n+1)}\left((t^n-1)+(t^n-t)+\cdots +(t^n-t^{n-1})\right)\\&=\dfrac{2(t-1)^2}{n(n+1)}\left((t^{n-1}+t^{n-2}+\cdots +1)+t(t^{n-2}+t^{n-3}+\cdots +1)+\cdots +t^{n-1}\right),\end{split}\]显然，在 $t>0$ 且 $t\ne 1$ 时恒有$$a_{n+1}-a_n>0,$$故 $a_{n+1}>a_n$．