Prove the construction of the polynomial function
$$f(x)=(1+x)^{2 n+1} \text {. }$$

Observe that, the right-hand side of the equation to be proved is obviously the coefficient of the $n$-th term of $f(x)$.
$$\begin{array}{l} \text { Also, } f(x)=(1+x)\left(1+2 x+x^{2}\right)^{n} \\ =(1+x) \sum_{k=0}^{n} \mathrm{C}_{n}^{k}\left(1+x^{2}\right)^{n-k}(2 x)^{k} \\ =\sum_{k=0}^{n} 2^{k} \mathrm{C}_{n}^{k}(1+x)\left(1+x^{2}\right)^{n-k} x^{k}, \end{array}$$

Then, when $n-k$ is even, the coefficient of $x^{n-k}$ in $(1+x)\left(1+x^{2}\right)^{n-k}$ is $\mathrm{C}_{n-k}^{\frac{n-k}{2}}$; when $n-k$ is odd, the coefficient of $x^{n-k}$ in $(1+x)\left(1+x^{2}\right)^{n-k}$ is $\mathrm{C}_{n-k}^{\frac{n-k-1}{2}}$.
Thus, for $k=0,1, \cdots, n$, in
$$2^{k} \mathrm{C}_{n}^{k}(1+x)\left(1+x^{2}\right)^{n-k} x^{k}$$

the coefficient of $x^{n}$ is always $2^{k} \mathrm{C}_{n}^{k} \mathrm{C}_{n-k}^{\left[\frac{n-k}{2}\right]}$.
Therefore, the coefficient of the $n$-th term of $f(x)$ is $\sum_{k=0}^{n} 2^{k} \mathrm{C}_{n}^{k} \mathrm{C}_{n-k}^{\left[\frac{n-k}{2}\right]}$.
In conclusion, $\sum_{k=0}^{n} 2^{k} \mathrm{C}_{n}^{k} \mathrm{C}_{n-k}^{\left[\frac{n-k}{2}\right]}=\mathrm{C}_{2 n+1}^{n}$.