已知数列 $\{a_n\}$ 满足 $a_1=2$，$a_{n+1}=\dfrac 19\left(a_n+2\sqrt{4a_n+1}+2\right)$．
1、求 $\{a_n\}$ 的通项公式．
2、数列 $\left\{\dfrac 1{a_n}\right\}$ 的前 $n$ 项和 $T_n$，求证：$T_n>\dfrac {4n}3-\dfrac {14}9$．
解析
1、根据题意，有\[a_{n+1}=\dfrac{a_n+4\sqrt{a_n+\dfrac 14}+2}{9},\]即\[\left(\sqrt{a_{n+1}+\dfrac 14}\right)^2=\dfrac{\left(\sqrt{a_n+\dfrac 14}+2\right)^2}9,\]于是\[\sqrt{a_{n+1}+\dfrac 14}=\dfrac 13\sqrt{a_n+\dfrac 14}+\dfrac 23,\]进而\[\sqrt{a_n+\dfrac 14}=1+\dfrac 32\cdot \dfrac{1}{3^n},\]于是\[a_n=\dfrac{3^{2n+1}+4\cdot 3^{n+1}+9}{4\cdot 3^{2n}},n\in\mathbb N^{\ast}.\]
2、根据题意，有\[ T_n=\sum_{k=1}^n\dfrac{4\cdot 3^{2k}}{3^{2k+1}+4\cdot 3^{k+1}+9}=\dfrac {4n}3-\sum_{k=1}^n\dfrac{16\cdot 3^k+12}{3^{2k+1}+4\cdot 3^{k+1}+9},\]于是原问题即证明\[\sum_{k=1}^n\dfrac{8\cdot 3^k+6}{3^{2k}+4\cdot 3^{k}+3}<\dfrac 73.\]而\[\begin{split} LHS&=\sum_{k=1}^{n}\left(\dfrac{3}{3^{k-1}+1}-\dfrac{1}{3^k+1}\right)\\ &=\dfrac 12-\dfrac{1}{3^n+1}+\sum_{k=1}^n\dfrac{2}{3^{k-1}+1}\\ &<2+\sum_{k=3}^n\dfrac{2}{3^{k-1}}\\ &<2+\dfrac{\dfrac 29}{1-\dfrac 13}=\frac 73,\end{split}\]于是命题得证．