Example 10 Find $\sum_{k=0}^{n} \frac{1}{2^{k}} C_{n+k}^{k}$.


Solution: Let the original expression be $f(n)$, then $f(0)=1$. By the identity (II), we have
$$
\begin{array}{l}
f(n+1)=\sum_{k=0}^{n+1} \mathrm{C}_{n+1+k}^{k} \cdot \frac{1}{2^{k}} \\
=\sum_{k=0}^{n+1} \mathrm{C}_{n+k}^{k} \cdot \frac{1}{2^{k}}+\sum_{k=1}^{n+1} \mathrm{C}_{n+k}^{k-1} \cdot \frac{1}{2^{k}} .
\end{array}
$$

Split the first term into $f(n)+\mathrm{C}_{2 n+1}^{n+1} \cdot \frac{1}{2^{n+1}}$. By changing the upper and lower indices of the combination number and the summation index in the second term, and substituting $k$ for the original $k-1$, we get
$$
\sum_{k=1}^{n+1} \mathrm{C}_{n+k}^{k-1} \cdot \frac{1}{2^{k}}=\sum_{k=0}^{n} \mathrm{C}_{n+k+1}^{k} \cdot \frac{1}{2^{k+1}} .
$$

Thus, $f(n+1)$
$$
=f(n)+\mathrm{C}_{2 n+1}^{n+1} \cdot \frac{1}{2^{n+1}}+\frac{1}{2} \sum_{k=0}^{n} \mathrm{C}_{n+k+1}^{k} \cdot \frac{1}{2^{k}} \text {. }
$$

Considering
$$
\begin{array}{l}
\mathrm{C}_{2 n+2}^{n+1}=\frac{(2 n+2)!}{(n+1)!(n+1)!} \\
=2 \cdot \frac{(2 n+1)!}{n!(n+1)!}=2 \mathrm{C}_{2 n+1}^{n+1},
\end{array}
$$

then $f(n+1)$
$$
\begin{array}{l}
=f(n)+\frac{1}{2}\left(\frac{1}{2^{n+1}} \mathrm{C}_{2 n+2}^{n+1}+\sum_{k=0}^{n} \mathrm{C}_{n+k+1}^{k} \cdot \frac{1}{2^{k}}\right) \\
=f(n)+\frac{1}{2} \sum_{k=0}^{n+1} \mathrm{C}_{n+k+1}^{k} \cdot \frac{1}{2^{k}} \\
=f(n)+\frac{1}{2} f(n+1) .
\end{array}
$$

Therefore, $f(n+1)=2 f(n)$.
Thus, $f(n)=2^{n}$.