Tính tổng $T=\frac{C_{2020}^0}{3}-\frac{C_{2020}^1}{4}+\frac{C_{2020}^2}{5}-\frac{X_{2020}^3}{6}+\mathrm{...}-\frac{C_{2020}^{2019}}{2022}+\frac{C_{2020}^{2020}}{2023}.$

Xét khai triển:

$x^2{\left(1-x\right)}^{2020}=x^2\sum _{k=0}^{2020}{C}_{2020}^k{\left(-x\right)}^k$

$=x^2\left(C_{2020}^0-C_{2020}^{1}x+C_{2020}^2{x}^2-C_{2020}^3{x}^3+\mathrm{...}-C_{2020}^{2019}{x}^{2019}+C_{2020}^{2020}{x}^{2020}\right)$

$=C_{2020}^0{x}^2-C_{2020}^1{x}^3+C_{2020}^2{x}^4-C_{2020}^3{x}^5+\mathrm{...}-C_{2020}^{2019}{x}^{2021}+C_{2020}^{2020}{x}^{2022}$

Lấy tích phân hai vế ta có:

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

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

$=\frac{1}{3}{C}_{2020}^0-\frac{1}{4}{C}_{2020}^1+\frac{1}{5}{C}_{2020}^2-\frac{1}{6}{C}_{2020}^3+\mathrm{...}-\frac{1}{2022}{C}_{2020}^{2019}+\frac{1}{2023}{C}_{2020}^{2020}$

Suy ra $T=\underset{0}{\overset{1}{\int }}{x}^2{\left(1-x\right)}^{2020}dx.$

Đặt $t=1-x\Rightarrow dt=-dx$. Đổi cận $\left\{\begin{array}{l}x=0\Rightarrow t=1\\ x=1\Rightarrow t=0\end{array}\right..$ Khi đó ta có:

$T=\underset{0}{\overset{1}{\int }}{x}^2{\left(1-x\right)}^{2020}dx=-\underset{1}{\overset{0}{\int }}{\left(1-t\right)}^2{t}^{2020}dt$

$=\underset{0}{\overset{1}{\int }}{t}^{2020}\left(t^2-2t+1\right)dt=\underset{0}{\overset{1}{\int }}\left(t^{2022}-2t^{2021}+t^{2020}\right)dt$

$=\left(\frac{t^{2023}}{2023}-2\frac{t^{2022}}{2022}+\frac{t^{2021}}{2021}\right)\left|\begin{array}{l}1\\ 0\end{array}\right.$

$=\frac{1}{2023}-\frac{2}{2022}+\frac{1}{2021}=\frac{1}{4133456313}$

Chọn C.