（4分）已知数列$\{a_{n}\}$满足$a_{1}=1$，$a_{n+1}=a_{n}-\dfrac{1}{3}a_{n}^{2}(n\in N^{*})$，则(　　)
A．$2 < 100a_{100} < \dfrac{5}{2}$              B．$\dfrac{5}{2} < 100a_{100} < 3$              C．$3 < 100a_{100} < \dfrac{7}{2}$              D．$\dfrac{7}{2} < 100a_{100} < 4$
分析：分析可知数列$\{a_{n}\}$是单调递减数列，根据题意先确定上限，得到${a}_{n}\leqslant \dfrac{3}{n+2}$，由此可推得$100a_{n} < 3$，再将原式变形确定下限，可得$\dfrac{1}{{a}_{n+1}}\leqslant \dfrac{1}{3}n+\dfrac{1}{3}(\dfrac{1}{2}+\dfrac{1}{3}+\ldots \ldots +\dfrac{1}{n+1})+1$，由此可推得$100{a}_{100} > \dfrac{5}{2}$，综合即可得到答案．
解：$\because a_{n+1}-a_{n}=-\dfrac{1}{3}a_{n}^{2} < 0$，
$\therefore \{a_{n}\}$为递减数列，
又${a}_{n+1}={a}_{n}-\dfrac{1}{3}{{a}_{n}}^{2}\leqslant \dfrac{2}{3}$，且$a_{n}\ne 0$，
$\therefore$$\dfrac{{a}_{n+1}}{{a}_{n}}=1-\dfrac{1}{3}{a}_{n}\geqslant \dfrac{2}{3} > 0$，
又$a_{1}=1 > 0$，则$a_{n} > 0$，
$\therefore$${a}_{n}-{a}_{n+1}=\dfrac{1}{3}{{a}_{n}}^{2}\geqslant \dfrac{1}{3}{a}_{n}{a}_{n+1}$，
$\therefore$$\dfrac{1}{{a}_{n+1}}-\dfrac{1}{{a}_{n}}\geqslant \dfrac{1}{3}$，
$\therefore$$\dfrac{1}{{a}_{n}}\geqslant \dfrac{1}{{a}_{1}}+\dfrac{1}{3}(n-1)=\dfrac{1}{3}n+\dfrac{2}{3}$，则${a}_{n}\leqslant \dfrac{3}{n+2}$，
$\therefore$$100{a}_{100}\leqslant 100\times \dfrac{3}{102} < \dfrac{306}{102}=3$；
由${a}_{n+1}={a}_{n}-\dfrac{1}{3}{{a}_{n}}^{2}$得${a}_{n+1}={a}_{n}(1-\dfrac{1}{3}{a}_{n})$，得$\dfrac{1}{{a}_{n+1}}-\dfrac{1}{{a}_{n}}=\dfrac{1}{3-{a}_{n}}\leqslant \dfrac{1}{3-\dfrac{3}{n+2}}=\dfrac{1}{3}(1+\dfrac{1}{n+1})$，
累加可得，$\dfrac{1}{{a}_{n+1}}\leqslant \dfrac{1}{3}n+\dfrac{1}{3}(\dfrac{1}{2}+\dfrac{1}{3}+\ldots \ldots +\dfrac{1}{n+1})+1$，
$\therefore$$\dfrac{1}{{a}_{100}}\leqslant 34+\dfrac{1}{3}\times (\dfrac{1}{2}+\dfrac{1}{3}+\ldots \ldots +\dfrac{1}{100}) < 34+\dfrac{1}{3}\times (\dfrac{1}{2}\times 6+\dfrac{1}{8}\times 93) < 40$，
$\therefore$$100{a}_{100} > 100\times \dfrac{1}{40}=\dfrac{5}{2}$；
综上，$\dfrac{5}{2} < 100{a}_{100} < 3$．
故选：$B$．
点评：本题考查递推数列，数列的单调性等知识，对化简变形能力要求较高，考查运算求解能力，逻辑推理能力，属于难题．