Lời giải chi tiết:

\(M = {2^{2016}}.C_{2017}^1 + {2^{2014}}.C_{2017}^3 + ... + {2^0}.C_{2017}^{2017}\)

\( + )\)Xét khai triển: \({\left( {2 + x} \right)^{2017}} = C_{2017}^0{.2^{2017}}.{x^0} + C_{2017}^1{.2^{2016}}.{x^1} + C_{2017}^2{.2^{2015}}.{x^2} + ... + C_{2017}^{2017}{.2^0}.{x^{2017}}\)

\( + )\)Thay \(x = 1\) vào hai vế:

\({\left( {2 + 1} \right)^{2017}} = C_{2017}^0{.2^{2017}} + C_{2017}^1{.2^{2016}} + C_{2017}^2{.2^{2015}} + ... + C_{2017}^{2017}{.2^0}\)  \(\left( 1 \right)\)

\( + )\)Thay \(x =  - 1\) vào hai vế:

\({\left( {2 - 1} \right)^{2017}} = C_{2017}^0{.2^{2017}} - C_{2017}^1{.2^{2016}} + C_{2017}^2{.2^{2015}} - ... - C_{2017}^{2017}{.2^0}\)  \(\left( 2 \right)\)

Lấy \(\left( 1 \right)\)trừ \(\left( 2 \right)\):

\({3^{2017}} - 1 = 2C_{2017}^1{.2^{2016}} + 2C_{2017}^3{.2^{2015}} + ... + 2C_{2017}^{2017}{.2^0}\)

\( \Leftrightarrow \dfrac{{{3^{2017}} - 1}}{2} = C_{2017}^1{.2^{2016}} + C_{2017}^3{.2^{2015}} + ... + C_{2017}^{2017}{.2^0}\)

\( \Leftrightarrow P = \dfrac{{{3^{2017}} - 1}}{2}\)

Chọn D.