Đáp án đúng: D Chọn đáp án D Ta có ${{\left( 1+x \right)}^{2n}}=C_{2n}^{0}+C_{2n}^{1}x+C_{2n}^{2}{{x}^{2}}+...+C_{2n}^{2n}{{x}^{n}}$ $\Rightarrow \left\{ \begin{array}{l}{{\left( 1+1 \right)}^{2n}}={{2}^{2n}}=C_{2n}^{0}+C_{2n}^{1}+C_{2n}^{2}+...+C_{2n}^{2n}\\{{\left( 1-1 \right)}^{2n}}=0=C_{2n}^{0}-C_{2n}^{1}+C_{2n}^{2}-...+C_{2n}^{2n}\end{array} \right.$ $\Rightarrow 2\left( C_{2n}^{0}+C_{2n}^{2}+...+C_{2n}^{2n} \right)={{2}^{2n}}+0\Rightarrow C_{2n}^{0}+C_{2n}^{2}+...+C_{2n}^{2n}={{2}^{2n-1}}$.