Cho \(A = \frac{1}{2}C_{2018}^1 + \frac{1}{4}C_{20...

Câu hỏi: Cho \(A = \frac{1}{2}C_{2018}^1 + \frac{1}{4}C_{2018}^3 + \frac{1}{6}C_{2018}^5 + ... + \frac{1}{{2018}}C_{2018}^{2017}\) . Ta có 2019A bằng:

A. \({2^{2018}} + 1\)

B. \({2^{2018}} - 1\)

C. \({2^{2019}} - 1\)

D. \({2^{2017}} + 1\)