19．（12分）设$\{a_{n}\}$是首项为1的等比数列，数列$\{b_{n}\}$满足$b_{n}=\dfrac{n{a}_{n}}{3}$，已知$a_{1}$，$3a_{2}$，$9a_{3}$成等差数列．
（1）求$\{a_{n}\}$和$\{b_{n}\}$的通项公式；
（2）记$S_{n}$和$T_{n}$分别为$\{a_{n}\}$和$\{b_{n}\}$的前$n$项和．证明：$T_{n}<\dfrac{{S}_{n}}{2}$．
分析：（1）根据$a_{1}$，$3a_{2}$，$9a_{3}$成等差数列，$\{a_{n}\}$是首项为1的等比数列，求出公比$q$，进一步求出$\{a_{n}\}$和$\{b_{n}\}$的通项公式；
（2）分别利用等比数列的前$n$项和公式和错位相减法，求出$S_{n}$和$T_{n}$，再利用作差法证明$T_{n}<\dfrac{{S}_{n}}{2}$．
解：（1）$\because a_{1}$，$3a_{2}$，$9a_{3}$成等差数列，$\therefore 6a_{2}=a_{1}+9a_{3}$，
$\because \{a_{n}\}$是首项为1的等比数列，设其公比为$q$，
则$6q=1+9q^{2}$，$\therefore q=\dfrac{1}{3}$，
$\therefore a_{n}=a_{1}q^{n-1}=(\dfrac{1}{3})^{n-1}$，
$\therefore b_{n}=\dfrac{n{a}_{n}}{3}=n\cdot (\dfrac{1}{3})^{n}$．
（2）证明：由（1）知$a_{n}=(\dfrac{1}{3})^{n-1}$，$b_{n}=n\cdot (\dfrac{1}{3})^{n}$，
$\therefore$${S}_{n}=\dfrac{1\times [1-(\dfrac{1}{3})^{n}]}{1-\dfrac{1}{3}}=\dfrac{3}{2}-\dfrac{1}{2}\times (\dfrac{1}{3})^{n-1}$，
${T}_{n}=1\times (\dfrac{1}{3})^{1}+2\times (\dfrac{1}{3})^{2}+\ldots +n\cdot (\dfrac{1}{3})^{n}$，①
$\therefore$$\dfrac{1}{3}{T}_{n}=1\times (\dfrac{1}{3})^{2}+2\times (\dfrac{1}{3})^{3}+\ldots +n\cdot (\dfrac{1}{3})^{n+1}$，②
①$-$②得，$\dfrac{2}{3}{T}_{n}=\dfrac{1}{2}[1-(\dfrac{1}{3})^{n}]-n(\dfrac{1}{3})^{n+1}$，
$\therefore$${T}_{n}=\dfrac{3}{4}-\dfrac{1}{4}\times (\dfrac{1}{3})^{n-1}-\dfrac{n}{2}(\dfrac{1}{3})^{n}$，
$\therefore$${T}_{n}-\dfrac{{S}_{n}}{2}=\dfrac{3}{4}-\dfrac{1}{4}\times (\dfrac{1}{3})^{n-1}-\dfrac{n}{2}\cdot (\dfrac{1}{3})^{n}-[\dfrac{3}{4}-\dfrac{1}{4}\times (\dfrac{1}{3})^{n-1}]<0$，
$\therefore T_{n}<\dfrac{{S}_{n}}{2}$．
点评：本题考查了等差数列与等比数列的性质，等比数列的前$n$项和公式和利用错位相减法求数列的前$n$项和，考查了方程思想和转化思想，属中档题．