A set contains $\left(2n+1\right)$ elements .the number of subsets of this set contains more then $n$ elements is equal to

The number of ways of selecting $n+1$ or more from the given set of $\left(2n+1\right)$ elements is 

${\left(2n+1\right)}_{C_{n+1}}+{\left(2n+1\right)}_{C_{n+2}}+...+{\left(2n+1\right)}_{C_{2n+1}}$

This is half of sum of the binomial coefficients of ${\left(1+x\right)}^{2n+1}$

Hence, the required value is $\frac{1}{2}\left(2^{2n+1}\right)=\boxed{{\mathbf{2}}^{\mathbf{2}\mathbf{n}}}$