Bài tập nâng cao · Bài 48
Bài tập nâng cao Các quy tắc tính đạo hàm · Bài 48
Giá trị của tổng $S=2.1C_{2021}^2+4.3C_{2021}^4+...2k(2k-1)C_{2021}^{2k}+...+2020.2019C_{2021}^{2020}$ bằng?
Xem lời giải
Lời giải
Xét biểu thức: $f(x)=(1+x)^{2021}=C_{2021}^0+C_{2021}^1x+C_{2021}^2x^2+C_{2021}^3x^3+...+C_{20201}^{2020}x^{2020}+C_{2021}^{2021}x^{2021}$
$f'(x)=2021(1+x)^{2020}=C_{2021}^1+2C_{2021}^2x+3C_{2021}^3x^2+...+2020C_{20201}^{2020}x^{2019}+2021C_{2021}^{2021}x^{2020}$
$f''(x)=2021.2020(1+x)^{2019}=2.1C_{2021}^2+3.2C_{2021}^3x+...+2020.2019C_{20201}^{2020}x^{2018}+2021.2020C_{2021}^{2021}x^{2019}$ $f''(1)=2021.2020.2^{2019}=2.1C_{2021}^2+3.2C_{2021}^3+...+2020.2019C_{20201}^{2020}+2021.2020C_{2021}^{2021}$
$f''(-1)=0=2.1C_{2021}^2-3.2C_{2021}^3+...+2020.2019C_{20201}^{2020}-2021.2020C_{2021}^{2021}$
$\begin{array}{l} f''(1)+f''(-1)=2021.2020.2^{2019}=2\left[ 2.1C_{2021}^2+4.3C_{2021}^4+...+2020.2019C_{20201}^{2020} \right] \\ \Leftrightarrow 2021.2020.2^{2018}=2.1C_{2021}^2+4.3C_{2021}^4+...+2020.2019C_{20201}^{2020} \end{array}$
Vậy $S=2021.2020.2^{2018}$
$f'(x)=2021(1+x)^{2020}=C_{2021}^1+2C_{2021}^2x+3C_{2021}^3x^2+...+2020C_{20201}^{2020}x^{2019}+2021C_{2021}^{2021}x^{2020}$
$f''(x)=2021.2020(1+x)^{2019}=2.1C_{2021}^2+3.2C_{2021}^3x+...+2020.2019C_{20201}^{2020}x^{2018}+2021.2020C_{2021}^{2021}x^{2019}$ $f''(1)=2021.2020.2^{2019}=2.1C_{2021}^2+3.2C_{2021}^3+...+2020.2019C_{20201}^{2020}+2021.2020C_{2021}^{2021}$
$f''(-1)=0=2.1C_{2021}^2-3.2C_{2021}^3+...+2020.2019C_{20201}^{2020}-2021.2020C_{2021}^{2021}$
$\begin{array}{l} f''(1)+f''(-1)=2021.2020.2^{2019}=2\left[ 2.1C_{2021}^2+4.3C_{2021}^4+...+2020.2019C_{20201}^{2020} \right] \\ \Leftrightarrow 2021.2020.2^{2018}=2.1C_{2021}^2+4.3C_{2021}^4+...+2020.2019C_{20201}^{2020} \end{array}$
Vậy $S=2021.2020.2^{2018}$