Tổng \(S = C_{2019}^0 + C_{2019}^3 + C_{2019}^6 + ... + C_{2019}^{2019}\) bằng

+ Ta tìm các số phức \(z\) thỏa mãn \({z^3} = 1\). Ta có \({z^3} = 1 \Leftrightarrow {z^3} - 1 = 0 \Leftrightarrow \left( {z - 1} \right)\left( {{z^2} + z + 1} \right) = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}z = 1\\{z^2} + z + 1 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{z_1} = 1\\{z_2} =  - \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i\\{z_3} =  - \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}i\end{array} \right.\)

+ Xét khai triển

\({\left( {1 + x} \right)^{2019}} = \sum\limits_{k = 0}^{2019} {C_{2019}^k{x^k} = C_{2019}^0 + C_{2019}^1x + C_{2019}^2{x^2} + C_{2019}^3{x^3} + C_{2019}^4{x^4} + C_{2019}^5{x^5} + C_{2019}^6{x^6} + ... + C_{2019}^{2019}{x^{2019}}} \) (*)

+ Thay \({z_2} =  - \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i\) vào khai triển (*) ta được

\(\begin{array}{l}{\left( {1 - \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i} \right)^{2019}} \\= C_{2019}^0 + C_{2019}^1{z_2} + C_{2019}^2{z_2}^2 + C_{2019}^3{z_2}^3 \\+ C_{2019}^4{z_2}^4 + C_{2019}^5{z_2}^5 + C_{2019}^6{z_2}^6 + ... + C_{2019}^{2019}{z_2}^{2019}\\ \Leftrightarrow {\left( {\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i} \right)^{2019}} = C_{2019}^0 + C_{2019}^1{z_2} + C_{2019}^2{z_2}^2\\ + C_{2019}^3 + C_{2019}^4{z_2} + C_{2019}^5{z_2}^2 + C_{2019}^6 + ... + C_{2019}^{2019}\\ \Leftrightarrow  - 1 = \left( {C_{2019}^0 + C_{2019}^3 + C_{2019}^6 + ... + C_{2019}^{2019}} \right) + {z_2}\left( {C_{2019}^1 + C_{2019}^4 + ... + C_{2019}^{2017}} \right) + z_2^2\left( {C_{2019}^2 + C_{2019}^5 + ... + C_{2019}^{2018}} \right)\,\left( 1 \right)\end{array}\)

+ Tương tự thay \({z_3} =  - \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}i\) vào khai triển (*) ta được

\(\begin{array}{l}{\left( {1 - \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}i} \right)^{2019}} = C_{2019}^0 + C_{2019}^1{z_3} + C_{2019}^2{z_3}^2 \\+ C_{2019}^3{z_3}^3 + C_{2019}^4{z_3}^4 + C_{2019}^5{z_3}^5 + C_{2019}^6{z_3}^6 + ... + C_{2019}^{2019}{z_3}^{2019}\\ \Leftrightarrow {\left( {\dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}i} \right)^{2019}} = C_{2019}^0 + C_{2019}^1{z_3} + C_{2019}^2{z_3}^2\\ + C_{2019}^3 + C_{2019}^4{z_3} + C_{2019}^5{z_3}^2 + C_{2019}^6 + ... + C_{2019}^{2019}\\ \Leftrightarrow  - 1 = \left( {C_{2019}^0 + C_{2019}^3 + C_{2019}^6 + ... + C_{2019}^{2019}} \right) + {z_3}\left( {C_{2019}^1 + C_{2019}^4 + ... + C_{2019}^{2017}} \right) + z_3^2\left( {C_{2019}^2 + C_{2019}^5 + ... + C_{2019}^{2018}} \right)\left( 2 \right)\end{array}\)

+ Thay \(z = 1\) vào vào khai triển (*) ta được

\(\begin{array}{l}{2^{2019}} = C_{2019}^0 + C_{2019}^1 + C_{2019}^2 + C_{2019}^3 + C_{2019}^4 + C_{2019}^5 + C_{2019}^6 + ... + C_{2019}^{2019}\\ \Leftrightarrow {2^{2019}} = \left( {C_{2019}^0 + C_{2019}^3 + C_{2019}^6 + ... + C_{2019}^{2019}} \right) + \left( {C_{2019}^1 + C_{2019}^4 + C_{2019}^7 + ... + C_{2019}^{2017}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {C_{2019}^2 + C_{2019}^5 + C_{2019}^8 + ... + C_{2019}^{2018}} \right)\left( 3 \right)\end{array}\)

Cộng vế với vế của (1) (2) và (3) ta được

\(\begin{array}{l}{2^{2019}} - 2 = 3\left( {C_{2019}^0 + C_{2019}^3 + C_{2019}^6 + ... + C_{2019}^{2019}} \right) + \left( {1 + {z_2} + {z_3}} \right)\left( {C_{2019}^1 + C_{2019}^4 + C_{2019}^7 + ... + C_{2019}^{2018}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {1 + z_2^2 + z_3^2} \right)\left( {C_{2019}^2 + C_{2019}^5 + C_{2019}^8 + ... + C_{2019}^{2017}} \right)\end{array}\)

Nhận thấy \(1 + {z_2} + {z_3} = 1 - \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i - \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}i = 0\) và \(1 + z_2^2 + z_3^2 = 1 + {\left( { - \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i} \right)^2} + {\left( { - \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}i} \right)^2} = 0\)

Nên \({2^{2019}} - 2 = 3S \Leftrightarrow S = \dfrac{{{2^{2019}} - 2}}{3}\)