Đáp án đúng: A
Giải chi tiết:\(\begin{array}{l}S = C_{2017}^2 + C_{2017}^3 + C_{2017}^4 + C_{2017}^5... + C_{2017}^{1008}\\ \Leftrightarrow S + C_{2017}^0 + C_{2017}^1 = C_{2017}^0 + C_{2017}^1 + C_{2017}^2 + C_{2017}^3 + C_{2017}^4 + C_{2017}^5... + C_{2017}^{1008}\\ \Leftrightarrow S + 2018 = C_{2017}^0 + C_{2017}^1 + C_{2017}^2 + C_{2017}^3 + C_{2017}^4 + C_{2017}^5... + C_{2017}^{1008}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\end{array}\)
Áp dụng tính chất \(C_n^k = C_n^{n - k}\) ta có:
\(C_{2017}^0 + C_{2017}^1 + C_{2017}^2 + C_{2017}^3 + C_{2017}^4 + C_{2017}^5... + C_{2017}^{1008} = C_{2017}^{2017} + C_{2017}^{2016} + C_{2017}^{2015} + C_{2017}^{2014} + C_{2017}^{2013} + C_{2017}^{2012}... + C_{2017}^{1009}\)
Suy ra \(S + 2018 = C_{2017}^{2017} + C_{2017}^{2016} + C_{2017}^{2015} + C_{2017}^{2014} + C_{2017}^{2013} + C_{2017}^{2012}... + C_{2017}^{1009}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
Cộng vế với vế của (1) và (2) ta có:
\(2S + 4036 = C_{2017}^0 + C_{2017}^1 + C_{2017}^2 + C_{2017}^3 + ... + C_{2017}^{1008} + C_{2017}^{1009} + ... + C_{2017}^{2016} + C_{2017}^{2017}\)
Ta có: \({\left( {a + b} \right)^{2017}} = C_{2017}^0{a^{2017}} + C_{2017}^1{a^{2016}}b + C_{2017}^2{a^{2015}}{b^2} + ... + C_{2017}^{2016}a{b^{2016}} + C_{2017}^{2017}{b^{2017}}\)
Thay \(a = 1,b = 1\) ta có:
\(\begin{array}{l}{2^{2017}} = C_{2017}^0 + C_{2017}^1 + C_{2017}^2 + ... + C_{2017}^{2016} + C_{2017}^{2017}\\ \Leftrightarrow {2^{2017}} = 2S + 4036 \Leftrightarrow {2^{2016}} = S + 2018 \Leftrightarrow S = {2^{2016}} - 2018\end{array}\)
Chọn A
