14.3.17 $\star$ Let $n>0$, find the greatest common divisor of $\mathrm{C}_{2 n}^{1}, \mathrm{C}_{2 n}^{3}, \cdots, \mathrm{C}_{2 n}^{2 n-1}$.
Let their greatest common divisor be $d$. Since $\mathrm{C}_{2 n}^{0}+\mathrm{C}_{2 n}^{1}+\mathrm{C}_{2 n}^{2}+\cdots+\mathrm{C}_{2 n}^{2 n}=2^{2 n}, \mathrm{C}_{2 n}^{0}-$ $\mathrm{C}_{2 n}^{1}+\mathrm{C}_{2 n}^{2}-\cdots+\mathrm{C}_{2 n}^{2 n}=0$, it follows that $\mathrm{C}_{2 n}^{1}+\mathrm{C}_{2 n}^{3}+\cdots+\mathrm{C}_{2 n}^{2 n-1}=2^{2 n-1}$, hence $d \mid 2^{2 n-1}$. Let $2^{k} \| n$. We will prove that $d=2^{k+1}$. Since $2^{k+1} \| \mathrm{C}_{2 n}^{1}$, it suffices to prove
$$
2^{k+1} \mid \mathrm{C}_{2 n}^{j}, j=3,5, \cdots, 2 n-1 .
$$

Since $j \mathrm{C}^{j}=2 n \mathrm{C}_{2 n-1}^{j-1}$, it follows that $2^{k+1} \mid j \mathrm{C}_{2 n}^{j}$. Also, $2 \nmid j$, so (1) holds. Therefore, $d=2^{k+1}$.