Áp dụng Nhị thức Newton, ta có $$(1+x)^{2019}=\mathrm{C}_{2019}^0+\mathrm{C}_{2019}^1x+\mathrm{C}_{2019}^2x^2+\cdots+\mathrm{C}_{2019}^{2019}x^{2019}$$

♥ Với \(x=-1\) ta có $$\begin{align*}
&(1-1)^{2019}=\mathrm{C}_{2019}^0-\mathrm{C}_{2019}^1+\mathrm{C}_{2019}^2-\cdots-\mathrm{C}_{2019}^{2019}\\
\Leftrightarrow&0=\mathrm{C}_{2019}^0-\mathrm{C}_{2019}^1+\mathrm{C}_{2019}^2-\cdots-\mathrm{C}_{2019}^{2019}\\
\Leftrightarrow&\mathrm{C}_{2019}^0+\mathrm{C}_{2019}^2+\cdots+\mathrm{C}_{2019}^{2018}=\mathrm{C}_{2019}^1+\mathrm{C}_{2019}^3+\cdots+\mathrm{C}_{2019}^{2019}\,\,(1)
\end{align*}$$

♥ Với \(x=1\) ta có $$\begin{align*}
&(1+1)^{2019}=\mathrm{C}_{2019}^0+\mathrm{C}_{2019}^1+\mathrm{C}_{2019}^2+\cdots+\mathrm{C}_{2019}^{2019}\\
\Leftrightarrow&2^{2019}=\mathrm{C}_{2019}^0+\mathrm{C}_{2019}^1+\mathrm{C}_{2019}^2+\cdots+\mathrm{C}_{2019}^{2019}\,\,(2)
\end{align*}$$

Từ (1) và (2) suy ra $$\begin{align*}
\mathrm{C}_{2019}^1+\mathrm{C}_{2019}^3+\cdots+\mathrm{C}_{2019}^{2019}&=\dfrac{2^{2019}}{2}\\
&=2^{2019-1}\\
&=2^{2018}
\end{align*}$$