91学 首页 > 数学 > 高考题 > 2021 > 2021年浙江 > 正文 返回 打印

2021年高考数学浙江10

  2022-05-03 08:24:59  

10.(4分)已知数列$\{a_{n}\}$满足$a_{1}=1$,$a_{n+1}=\dfrac{{a}_{n}}{1+\sqrt{{a}_{n}}}(n\in N*)$.记数列$\{a_{n}\}$的前$n$项和为$S_{n}$,则(  )
A.$\dfrac{3}{2}<S_{100}<3$              B.$3<S_{100}<4$              C.$4<S_{100}<\dfrac{9}{2}$              D.$\dfrac{9}{2}<S_{100}<5$
分析:由题意首先整理所给的递推关系式,得到数列的通项的范围,然后结合求和公式裂项即可确定前100项和的范围.
$\dfrac{1}{\sqrt{a_{n}}}-\dfrac{1}{\sqrt{a_{1}}}=\dfrac{1}{2}(n-1)\Rightarrow \dfrac{1}{\sqrt{a_{n}}}=1+\dfrac{n-1}{2}=\dfrac{n+1}{2}\Rightarrow a_{n}=\dfrac{4}{(n+1)^{2}}$
解:因为$a_{1}=1,a_{n+1}=\dfrac{a_{n}}{1+\sqrt{a_{n}}}$,所以$a_{n}>0,a_{2}=\dfrac{1}{2}$,所以$S_{100}>a_{1}+a_{2}=\dfrac{3}{2}$,
$\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}}+\dfrac{1}{\sqrt{a_{n}}}=(\dfrac{1}{\sqrt{a_{n}}}+\dfrac{1}{2})^{2}-\dfrac{1}{4}<(\dfrac{1}{\sqrt{a_{n}}}+\dfrac{1}{2})^{2}$,
$\therefore$$\dfrac{1}{\sqrt{a_{n+1}}}<\dfrac{1}{\sqrt{a_{n}}}+\dfrac{1}{2}$,故$\dfrac{1}{\sqrt{a_{n}}}-\dfrac{1}{\sqrt{a_{n-1}}}<\dfrac{1}{2},\cdots ,\dfrac{1}{\sqrt{a_{2}}}-\dfrac{1}{\sqrt{a_{1}}}<\dfrac{1}{2}$,
由累加法可得当$n\geqslant 2$ 时,
$\dfrac{1}{\sqrt{{a}_{n}?}}?\dfrac{1}{\sqrt{{a}_{1}?}}<\dfrac{1}{2}(n?1)\Rightarrow \dfrac{1}{\sqrt{{a}_{n}?}}<1+\dfrac{n?1}{2}=\dfrac{n+1}{2}\Rightarrow {a}_{n}>\dfrac{4}{{(n+1)}^{2}}$,
又因为当$n=1$ 时,$a_{n}=\dfrac{4}{(n+1)^{2}}$ 也成立,所以${a}_{n}\geqslant \dfrac{4}{{(n+1)}^{2}}(n\in {N}^{*})$,
所以$a_{n+1}=\dfrac{a_{n}}{1+\sqrt{a_{n}}}\leqslant \dfrac{a_{n}}{1+\dfrac{2}{n+1}}=\dfrac{n+1}{n+3}a_{n}$,
$\therefore$$\dfrac{a_{n+1}}{a_{n}}=\dfrac{n+1}{n+3}$,故$\dfrac{{a}_{n}}{{a}_{n-1}}\leqslant \dfrac{n}{n+2},\dfrac{{a}_{n-1}}{{a}_{n-2}}=\dfrac{n-1}{n+1},\cdots ,\dfrac{{a}_{2}}{{a}_{1}}=\dfrac{2}{4}$,
由累乘法可得当$n\geqslant 2$ 时,$a_{n}=\dfrac{a_{n}}{a_{1}}=\dfrac{n}{n+2}\times \dfrac{n-1}{n+1}\times \dfrac{n-2}{n}\times \cdots \times \dfrac{3}{5}\times \dfrac{2}{4}=\dfrac{6}{(n+2)(n+1)}=6(\dfrac{1}{n+1}-\dfrac{1}{n+2})$,
所以$S_{100}=1+6(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\cdots +\dfrac{1}{101}-\dfrac{1}{102})<1+6(\dfrac{1}{3}-\dfrac{1}{102})<1+2=3$.
故选:$A$.
点评:本题主要考查数列的递推关系式及其应用,数列求和与放缩的技巧等知识,属于难题.

http://x.91apu.com//shuxue/gkt/2021/2021zj/2022-05-03/33342.html