Chọn C

Ta có ${(1-x)}^{2020}=\sum _{k=0}^{2020}{C}_{2020}^k{(-x)}^k$ 

Xét hàm số:

$f\left(x\right)=x^2{(1-x)}^{2020}=\sum _{k=0}^{2020}{C}_{2020}^k{x}^{k+2}.{(-1)}^k=C_{2020}^0{x}^2-C_{2020}^1{x}^3+C_{2020}^2{x}^4-C_{2020}^3{x}^5+\mathrm{...}+C_{2020}^{2019}{x}^{2021}-C_{2020}^0{x}^{2022}$ 

Là hàm số liên tục trên $\mathrm{ℝ}$ nên:

$\underset{0}{\overset{1}{\int }}{x}^2{(1-x)}^{2020}dx=\underset{0}{\overset{1}{\int }}(C_{2020}^0{x}^2-C_{2020}^1{x}^3+C_{2020}^2{x}^4-C_{2020}^3{x}^5+\mathrm{...}+C_{2020}^{2019}{x}^{2021}-C_{2020}^0{x}^{2022})dx$ 

Ta xét

$VT=\underset{0}{\overset{1}{\int }}{x}^2{(1-x)}^{2020}dx$   (đặt $u=1-x\Rightarrow x=1-u\Rightarrow dx=-du;$ đổi biến $x=0\Rightarrow u=1;x=1\Rightarrow u=0)$ 

$\begin{array}{l}=\underset{1}{\overset{0}{\int }}{(1-u)}^2{u}^{2020}(-du)=\underset{0}{\overset{1}{\int }}(u^{2020}-2u^{2021}+u^{2022})du={\left.\frac{u^{2021}}{2021}-\frac{2u^{2022}}{2022}+\frac{u^{2023}}{2023}\right|}_0^1\\ =\frac{1}{2021}-\frac{2}{2022}+\frac{1}{2023}\\ =\frac{1}{4133456313}\end{array}$ 

$\begin{array}{l}VP=\underset{0}{\overset{1}{\int }}(C_{2020}^0{x}^2-C_{2020}^1{x}^3+C_{2020}^2{x}^4-C_{2020}^3{x}^5+\mathrm{...}-C_{2020}^{2019}{x}^{2021}+C_{2020}^0{x}^{2022})dx\\ ={\left.C_{2020}^0.\frac{x^3}{3}-C_{2020}^1.\frac{x^4}{4}+C_{2020}^2.\frac{x^5}{5}-C_{2020}^3.\frac{x^6}{6}+\mathrm{...}-C_{2020}^{2019}.\frac{x^{2022}}{2022}+C_{2020}^{2020}.\frac{x^{2023}}{2023}\right|}_0^1\\ =\frac{C_{2020}^0}{3}-\frac{C_{2020}^1}{4}+\frac{C_{2020}^2}{5}-\frac{C_{2020}^3}{6}+\mathrm{....}-\frac{C_{2020}^{2019}}{2022}+\frac{C_{2020}^{2020}}{2023}.\end{array}$ 

$VT=VP\Rightarrow \frac{C_{2020}^0}{3}-\frac{C_{2020}^1}{4}+\frac{C_{2020}^2}{5}-\frac{C_{2020}^3}{6}+\mathrm{....}-\frac{C_{2020}^{2019}}{2022}+\frac{C_{2020}^{2020}}{2023}=\frac{1}{4133456313}$.