Đáp án đúng: B
Giải chi tiết:Ta có \(C_{2017}^{2k} = C_{2017}^{2017 - 2k},\,\,\forall k \in \left\{ {0,1,....,1008} \right\}.\)
Do đó
\(\begin{array}{l}2T = \left( {C_{2017}^1 + C_{2017}^3 + C_{2017}^5 + ... + C_{2017}^{2017}} \right) + \left( {C_{2017}^{2016} + C_{2017}^{2014} + C_{2017}^{2012} + ... + C_{2017}^0} \right)\\\,\,\,\,\,\,\, = C_{2017}^0 + C_{2017}^1 + C_{2017}^2 + C_{2017}^3 + ... + C_{2017}^{2016} + C_{2017}^{2017}\\\,\,\,\,\,\,\, = C_{2017}^0{.1^0}{.1^{2017 - 0}} + C_{2017}^1{.1^1}{.1^{2017 - 1}} + C_{2017}^2{.1^2}{.1^{2017 - 2}} + C_{2017}^3{.1^3}{.1^{2017 - 3}} + ... + C_{2017}^{2016}{.1^{2016}}{.1^{2017 - 2016}} + C_{2017}^{2017}{.1^{2017}}{.1^{2017 - 2017}}\\\,\,\,\,\,\,\, = {\left( {1 + 1} \right)^{2017}} = {2^{2017}}.\end{array}\)
Do đó \(T = {2^{2016}}.\)
Chọn đáp án B.
