Phương pháp giải:

Sử dụng công thức khai triển Newton dạng: \({\left( {x + 1} \right)^n} = \sum\limits_{i = 0}^n {C_n^i{x^n}} \).

Lời giải chi tiết:

Ta có: \({\left( {x + 1} \right)^{2017}} = \sum\limits_{i = 0}^{2017} {C_{2017}^i{x^{2017}}} \)

\(\begin{array}{l}x = 1 \Rightarrow {\left( {1 + 1} \right)^{2017}} = \sum\limits_{i = 0}^{2017} {C_{2017}^i{1^{2017}}}  \Leftrightarrow C_{2017}^0 + C_{2017}^1 + C_{2017}^2 + C_{2017}^3 + ... + C_{2017}^{2016} + C_{2017}^{2017} = {2^{2017}}\\x =  - 1 \Rightarrow {\left( { - 1 + 1} \right)^{2017}} = \sum\limits_{i = 0}^{2017} {C_{2017}^i{{\left( { - 1} \right)}^{2017}}}  \Leftrightarrow C_{2017}^0 - C_{2017}^1 + C_{2017}^2 - C_{2017}^3 + ... + C_{2017}^{2016} - C_{2017}^{2017} = 0\\ \Rightarrow 2\left( {C_{2017}^1 + C_{2017}^3 + ... + C_{2017}^{2015} + C_{2017}^{2017}} \right) = {2^{2017}} - 0 \Leftrightarrow C_{2017}^1 + C_{2017}^3 + ... + C_{2017}^{2015} + C_{2017}^{2017} = {2^{2016}}\end{array}\)

Chọn: A