Xét khai triển $(1+x)^{2019}=C^0_{2019}+xC^1_{2019}+x^2C^2_{2019}+x^3C^3_{2019}...+x^{2019}C^{2019}_{2019}$Đạo hàm 2 vế, ta được
$2019(1+x)^{2018}=C^1_{2019}+2xC^2_{2019}+3x^2C^3_{2019}+...+2019x^{2018}C^{2019}_{2019}$
Thế $x=1$,ta có:
$S=C^1_{2019}+2C^2_{2019}+3C^3_{2019}+...+2019C^{2019}_{2019}=2019.2^{2018}$