The right hand side should be $C_{2n+1}^n$. Suppose we have numbers from $1$ to $2n+1$. We can use the following strategy to pick $n$ numbers out of these $2n+1$ numbers.
1) We split the numbers set into $n$ pairs and one singleton
$(1,2),(3,4),\cdots,(2n-1,2n),(2n+1)$.
2) We first pick $k$ pairs and then pick one number from each pair. So we get $k$ numbers now.
3) We then pick $[\frac{n-k}2]$ pairs from the rest $n-k$ pairs and both numbers in each pair will be picked. So far we have got $k+2[\frac{n-k}2]$ numbers.
4) If $n-k$ is even, we already got $n$ numbers; If $n-k$ is odd, we only had $n-1$ numbers and will pick $2n+1$.
Now it is easy to see the equation should hold.