Let the given expression be $a_n$, then $$\begin{align*} a_{n+1}=\ &\sum_{k=0}^{n+1}\frac{1}{2^k}\binom{n+1+k}{n+1} \\ =\ &\sum_{k=0}^{n+1}\frac{1}{2^k}\left(\binom{n+k}{n} +\binom{n+k}{n+1}\right) \\=\ &\sum_{k=0}^{n+1}\frac{1}{2^k}\binom{n+k}{n}+\sum_{k=0}^{n+1}\frac{1}{2^k}\binom{n+k}{n+1} \\ =\ & a_n + \frac{1}{2^{n+1}}\binom{2n+1}{n}+\frac{1}{2}\sum_{k=0}^{n}\frac{1}{2^k}\binom{n+k+1}{n+1} \\=\ & a_n + \frac{1}{2^{n+1}}\binom{2n+1}{n} + \frac{1}{2}a_{n+1}-\frac{1}{2^{n+2}}\binom{2n+2}{n+1} \\ =\ & a_n +\frac{1}{2}a_{n+1} +\frac{1}{2^{n+2}}\left(\frac{2\cdot(2n+1)!}{n!(n+1)!}-\frac{(2n+2)!}{(n+1)!^2}\right)\\=\ & a_n + \frac{1}{2}a_{n+1}\end{align*}$$

Thus, $$a_{n+1}=a_n+\frac{1}{2}a_{n+1}\implies a_{n+1}=2a_n$$

Because $a_0=1$, therefore $a_n =\boxed{2^n}$.