Tổng \(C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ... + C_{2019}^{1009}\) bằng:

Đáp án C

Phương pháp:

Áp dụng \(C_n^0 + C_n^1 + ... + C_n^n = {2^n}\).

Cách giải:

Ta có: \(C_{2019}^0 + C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ... + C_{2019}^{2019} = {2^{2019}}\)

Mà \(C_{2019}^0 = C_{2019}^{2019} = 1,\,\,C_{2019}^1 = C_{2019}^{2018},\,C_{2019}^2 = C_{2019}^{2017},...,C_{2019}^{1009} = C_{2019}^{1010}\)

\( \Leftrightarrow 1 + C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ... + C_{2019}^{1009} + C_{2019}^{1010} + ... + C_{2019}^1 + 1 = {2^{2019}}\)

\( \Leftrightarrow 2 + 2\left( {C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ... + C_{2019}^{1009}} \right) = {2^{2019}}\)

\( \Leftrightarrow C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ... + C_{2019}^{1009} = {2^{2018}} - 1\).