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．