# How can I express $\sum_{k=0}^n\binom{-1/2}{k}(-1)^k\binom{-1/2}{n-k}$ without using summations or minus signs?

How can I express $$\sum_{k=0}^n\binom{-1/2}{k}(-1)^k\binom{-1/2}{n-k}$$ without using summations or minus signs?

Here's a generating function approach. We have, by the binomial theorem, $$\sum_k \binom{-1/2}{k} (-1)^k z^k = (1-z)^{-1/2} \tag{1},$$ and $$\sum_k \binom{-1/2}{k} z^k = (1+z)^{-1/2}.$$ Since your sum is the convolution of $$\binom{-1/2}{k} (-1)^k$$ and $$\binom{-1/2}{k}$$, the generating function for your sum is $$(1-z)^{-1/2} (1+z)^{-1/2} = (1-z^2)^{-1/2}$$. Applying the binomial theorem again (Eq. $$(1)$$ with $$z^2$$ in place of $$z$$), we have $$\sum_{k=0}^n \binom{-1/2}{k} (-1)^k \binom{-1/2}{n-k} = \begin{cases} \binom{-1/2}{n/2} (-1)^{n/2}, & n \text{ is even;} \\ 0, & n \text{ is odd.} \end{cases}$$
But since $$\binom{-1/2}{n} = \left(\frac{-1}{4}\right)^n \binom{2n}{n}$$ (see Concrete Mathematics, p. 186, Eq. 5.37), this simplifies to $$\sum_{k=0}^n \binom{-1/2}{k} (-1)^k \binom{-1/2}{n-k} = \begin{cases} \frac{1}{2^n} \binom{n}{n/2}, & n \text{ is even;} \\ 0, & n \text{ is odd.} \end{cases}$$