Đáp án:
\[{2^{2018}}\]
Giải thích các bước giải:
Áp dụng \(C_n^k = C_n^{n - k}\) ta có:
\(\left\{ \begin{array}{l}
C_{2019}^1 = C_{2019}^{2018}\\
C_{2019}^3 = C_{2019}^{2016}\\
C_{2019}^5 = C_{2019}^{2014}\\
.....\\
C_{2019}^{2019} = C_{2019}^0
\end{array} \right.\)
Do đó,
\[\begin{array}{l}
A = C_{2019}^1 + C_{2019}^3 + C_{2019}^5 + .... + C_{2019}^{2019}\\
\Rightarrow 2A = \left( {C_{2019}^1 + C_{2019}^{2018}} \right) + \left( {C_{2019}^3 + C_{2019}^{2016}} \right) + \left( {C_{2019}^5 + C_{2019}^{2014}} \right) + .... + \left( {C_{2019}^{2019} + C_{2019}^0} \right)\\
\Leftrightarrow 2A = C_{2019}^0 + C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + .... + C_{2019}^{2018} + C_{2019}^{2019}\\
\Rightarrow 2A = {\left( {1 + 1} \right)^{2019}}\\
\Rightarrow 2A = {2^{2019}}\\
\Rightarrow A = {2^{2018}}
\end{array}\]
