Cho khai triển \({\left( {\sqrt 3  + x} \right)^{2019}} = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ..... + {a_{2019}}{x^{2019}}\).

\(\begin{array}{l}
{\left( {\sqrt 3  + x} \right)^{2019}} = \sum\limits_{k = 0}^{2019} {C_{2019}^k} {\left( {\sqrt 3 } \right)^k}{x^{2019 - k}}\\
 = C_{2019}^0{\left( {\sqrt 3 } \right)^{2019}} + C_{2019}^1{\left( {\sqrt 3 } \right)^{2018}}x + C_{2019}^2{\left( {\sqrt 3 } \right)^{2017}}{x^2} + ... + C_{2019}^{2018}.\sqrt 3 {x^{2018}} + C_{2019}^{2019}{x^{2019}}\\
 = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ... + {a_{2019}}{x^{2019}}
\end{array}\)

Ta có \({i^m} = \left\{ \begin{array}{l}
1\,\,\,\,\,\,\,\,khi\,\,\,m = 4l\\
i\,\,\,\,\,\,\,\,khi\,\,\,m = 4l + 1\\
 - 1\,\,\,\,khi\,\,\,m = 4l + 2\\
 - i\,\,\,\,khi\,\,\,m = 4l + 3
\end{array} \right.\left( {l \in Z} \right)\)

Chọn \(x =  - i\) ta có:

\(\begin{array}{l}
{\left( {\sqrt 3  - i} \right)^{2019}} = \sum\limits_{k = 0}^{2019} {C_{2019}^k{{\left( {\sqrt 3 } \right)}^k}{{\left( { - i} \right)}^{2019 - k}}} \\
 = C_{2019}^0{\left( {\sqrt 3 } \right)^{2019}} - C_{2019}^1{\left( {\sqrt 3 } \right)^{2018}}i + C_{2019}^2{\left( {\sqrt 3 } \right)^{2017}}{i^2} - ... + C_{2019}^{2018}.\sqrt 3 .{i^{2018}} - C_{2019}^{2019}{i^{2019}}\\
 = {a_0} - {a_1}i + {a_2}{i^2} - {a_3}{i^3} + ... + {a_{2018}}{i^{2018}} - {a_{2019}}{i^{2019}}\\
 = {a_0} - {a_1}i - {a_2} + {a_3}i + ... - {a_{2018}} + {a_{2019}}i\\
 \Rightarrow {\left( {\sqrt 3  + 1} \right)^{2019}} + {\left( {\sqrt 3  - 1} \right)^{2019}} = 2\left( {{a_0} - {a_2} + {a_4} - {a_6} + ... + {a_{2016}} - {a_{2018}}} \right)\\
 \Leftrightarrow 2S = {\left[ {{{\left( {\sqrt 3  + 1} \right)}^3}} \right]^{673}} + {\left[ {{{\left( {\sqrt 3  - 1} \right)}^3}} \right]^{673}} = {\left( {8i} \right)^{673}} + {\left( { - 8i} \right)^{673}} = 0\\
 \Leftrightarrow 2S = {8^{673}}.{i^{673}} - {8^{673}}.{i^{673}} = 0 \Leftrightarrow S = 0
\end{array}\)