2. Ta có
$2^{2n} =(1+1)^{2n} =\sum_{k=0}^{2n}C^k_{2n}=\left ( C^0_{2n}+C^2_{2n}+\ldots+C^{2n}_{2n} \right )+\left ( C^1_{2n}+C^3_{2n}+\ldots+C^{2n-1}_{2n} \right )$
$0 =(1-1)^{2n} =\sum_{k=0}^{2n}(-1)^kC^k_{2n}=\left ( C^0_{2n}+C^2_{2n}+\ldots+C^{2n}_{2n} \right )-\left ( C^1_{2n}+C^3_{2n}+\ldots+C^{2n-1}_{2n} \right )$
Suy ra
$\left ( C^0_{2n}+C^2_{2n}+\ldots+C^{2n}_{2n} \right )=\left ( C^1_{2n}+C^3_{2n}+\ldots+C^{2n-1}_{2n} \right )=\frac{2^{2n}}{2}=2^{2n-1}.$
Áp dụng vào bài toán ta suy ra $2^{2n-1}=2^{23}\Rightarrow n=12.$