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Phương pháp:
- Xét khai triển ${{x}^{2}}{{\left( 1-x \right)}^{2020}}.$
- Lấy tích phân từ 0 đến 1 hai vế, chứng minh $T=\int\limits_{0}^{1}{{{x}^{2}}{{\left( 1-x \right)}^{2020}}dx.}$
- Tính tích phân bằng phương pháp đổi biến số, đặt $t=1-x.$
Cách giải:
Xét khai triển:
${{x}^{2}}{{\left( 1-x \right)}^{2020}}={{x}^{2}}\sum\limits_{k=0}^{2020}{C_{2020}^{k}{{\left( -x \right)}^{k}}}$
$={{x}^{2}}\left( C_{2020}^{0}-C_{2020}^{1}x+C_{2020}^{2}{{x}^{2}}-C_{2020}^{3}{{x}^{3}}+...-C_{2020}^{2019}{{x}^{2019}}+C_{2020}^{2020}{{x}^{2020}} \right)$
$=C_{2020}^{0}{{x}^{2}}-C_{2020}^{1}{{x}^{3}}+C_{2020}^{2}{{x}^{4}}-C_{2020}^{3}{{x}^{5}}+...-C_{2020}^{2019}{{x}^{2021}}+C_{2020}^{2020}{{x}^{2022}}$
Lấy tích phân hai vế ta có:
$\int\limits_{0}^{1}{{{x}^{2}}{{\left( 1-x \right)}^{2020}}dx=}\int\limits_{0}^{1}{\left( C_{2020}^{0}{{x}^{2}}-C_{2020}^{1}{{x}^{3}}+C_{2020}^{2}{{x}^{4}}-C_{2020}^{3}{{x}^{5}}+...-C_{2020}^{2019}{{x}^{2021}}+C_{2020}^{2020}{{x}^{2022}} \right)dx}$
$=\left( C_{2020}^{0}\dfrac{{{x}^{3}}}{3}-C_{2020}^{1}\dfrac{{{x}^{4}}}{4}+C_{2020}^{2}\dfrac{{{x}^{5}}}{5}-C_{2020}^{3}\dfrac{{{x}^{6}}}{6}+...-C_{2020}^{2019}\dfrac{{{x}^{2022}}}{2022}+C_{2020}^{2020}\dfrac{{{x}^{2023}}}{2023} \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.$
$=\dfrac{1}{3}C_{2020}^{0}-\dfrac{1}{4}C_{2020}^{1}+\dfrac{1}{5}C_{2020}^{2}-\dfrac{1}{6}C_{2020}^{3}+...-\dfrac{1}{2022}C_{2020}^{2019}+\dfrac{1}{2023}C_{2020}^{2020}$
Suy ra $T=\int\limits_{0}^{1}{{{x}^{2}}{{\left( 1-x \right)}^{2020}}dx.}$
Đặt $t=1-x\Rightarrow dt=-dx$. Đổi cận $\left\{ \begin{aligned}
& x=0\Rightarrow t=1 \\
& x=1\Rightarrow t=0 \\
\end{aligned} \right..$ Khi đó ta có:
$T=\int\limits_{0}^{1}{{{x}^{2}}{{\left( 1-x \right)}^{2020}}dx}=-\int\limits_{1}^{0}{{{\left( 1-t \right)}^{2}}{{t}^{2020}}dt}$
$=\int\limits_{0}^{1}{{{t}^{2020}}\left( {{t}^{2}}-2t+1 \right)dt}=\int\limits_{0}^{1}{\left( {{t}^{2022}}-2{{t}^{2021}}+{{t}^{2020}} \right)dt}$
$=\left( \dfrac{{{t}^{2023}}}{2023}-2\dfrac{{{t}^{2022}}}{2022}+\dfrac{{{t}^{2021}}}{2021} \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.$
$=\dfrac{1}{2023}-\dfrac{2}{2022}+\dfrac{1}{2021}=\dfrac{1}{4133456313}$