Ta có:
$C_{2n+1}^1=C_{2n+1}^{2n}, C_{2n+1}^2=C_{2n+1}^{2n-1},\dots, C_{2n+1}^n=C_{2n+1}^{n+1}$. Suy ra
$2\left ( C^{1}_{2n+1}+C^{2}_{2n+1}+...+C^{n}_{2n+1} \right ) = C^{1}_{2n+1}+C^{2}_{2n+1}+...+C^{n}_{2n+1}+ C_{2n+1}^{n+1}+ \dots +C_{2n+1}^{2n}=\sum_{k=0}^{2n+1}C_{2n+1}^k- C_{2n+1}^0-C_{2n+1}^{2n+1}=2^{2n+1}-1-1=2^{2n+1}-2.$
Do đó
$2\left ( 2^{20}-1 \right )=2^{2n+1}-2\Rightarrow n=10.$
Tìm được $n$ thì đây là bài toán tổ hợp quen thuộc.