2022年高考数学浙江9<-->2022年高考数学浙江11
(4分)已知数列{an}满足a1=1,an+1=an−13a2n(n∈N∗),则( ) A.2<100a100<52 B.52<100a100<3 C.3<100a100<72 D.72<100a100<4 分析:分析可知数列{an}是单调递减数列,根据题意先确定上限,得到an⩽3n+2,由此可推得100an<3,再将原式变形确定下限,可得1an+1⩽13n+13(12+13+……+1n+1)+1,由此可推得100a100>52,综合即可得到答案. 解:∵, \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}, \therefore100{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, \therefore100{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.
2022年高考数学浙江9<-->2022年高考数学浙江11
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