Đề bài
Tính tổng sau đây:
\(C_{2021}^0 - 2C_{2021}^1 + {2^2}C_{2021}^2 - {2^3}C_{2021}^3 + ... - {2^{2021}}C_{2021}^{2021}\)
Lời giải chi tiết
Ta có:
\(\begin{array}{l}{(x - 2)^{2021}} = C_{2021}^0{x^{2021}} + C_{2021}^1{x^{2020}}( - 2) + C_{2021}^2{x^{2019}}{( - 2)^2} + C_{2021}^3{x^{2018}}{( - 2)^3} + ... + C_{2021}^{2021}{( - 2)^{2021}}\\ = C_{2021}^0{x^{2021}} - 2C_{2021}^1{x^{2020}} + {2^2}C_{2021}^2{x^{2019}} - {2^3}C_{2021}^3{x^{2018}} + ... - {2^{2021}}C_{2021}^{2021}\end{array}\)
Thay \(x = 1\) vào cả hai vế, ta suy ra
\(C_{2021}^0 - 2C_{2021}^1 + {2^2}C_{2021}^2 - {2^3}C_{2021}^3 + ... - {2^{2021}}C_{2021}^{2021} = {( - 1)^{2021}} = - 1\)