数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=\dfrac{4}{3}$，$a_{n+1}-1=a_{n}^{2}-a_{n}$（$n \in\mathbb N^{*}$），数列 $\left\{\dfrac{1}{a_{n}}\right\}$ 的前 $n$ 项和为 $S_{n}$，则（ ）
A．$1<S_{2021}<2$
B．$2<S_{2021}<3$
C．$3<S_{2021}<4$
D．$4<S_{2021}<5$
答案 B．
解析 根据题意，有\[\dfrac{1}{a_{n+1}-1}=\dfrac{1}{a_n-1}-\dfrac{1}{a_n},\]因此\[S_n=\sum_{k=1}^n\left(\dfrac{1}{a_k-1}-\dfrac{1}{a_{k+1}-1}\right)=\dfrac{1}{a_1-1}-\dfrac{1}{a_{n+1}-1}=3-\dfrac{1}{a_{n+1}-1}.\]根据题意，有\[a_{n+1}-a_n=(a_n-1)^2\geqslant \dfrac 19,\]因此\[a_{2022}\geqslant a_1+\dfrac{2021}9>2,\]因此 $\dfrac{1}{a_{2022}-1}\in (0,1)$，$S_{2021}\in (2,3)$．