Tổng \(C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ......

C

- Hướng dẫn giải

Phương pháp giải:

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

Giải chi tiết:

Ta có: \(C_{2019}^0 + C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ... + C_{2019}^{2009} = {2^{2019}}\)

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

\(\begin{array}{l} \Leftrightarrow 1 + C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ... + C_{2019}^{1009} + C_{2019}^{1010} + ... + C_{2019}^1 + 1 = {2^{2019}}\\ \Leftrightarrow 2 + 2\left( {C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ... + C_{2019}^{1009}} \right) = {2^{2019}}\\ \Leftrightarrow C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + ... + C_{2019}^{1009} = {2^{2018}} - 1.\end{array}\)

Chọn: C