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2022年高考数学浙江10

  2022-12-16 20:48:20  

(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.

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