Xét khai triển
$(x+1)^{2019} =C_{2019}^0 + xC_{2019}^1 + x^2 C_{2019}^2 +x^3C_{2019}^3+ ... + x^{2019} C_{2019}^{2019}$
Đạo hàm 2 vế ta có
$2019.(x+1)^{2018} =C_{2019}^1 + 2x C_{2019}^2 +3x^2 C_{2019}^3+ ... + 2019 x^{2018} C_{2019}^{2019}$
Thay $x=1$ ta được
$2019.2^{2018} = C_{2019}^1 + 2 C_{2019}^2 +3C_{2019}^3 ... + 2019 C_{2019}^{2019} =S$