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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?

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  • $\begingroup$ For a start, it's $0$ when $n$ is odd... $\endgroup$ – Henning Makholm Nov 8 '11 at 4:05
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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 $$\sum_{k=0}^n \binom{-1/2}{k} (-1)^k \binom{-1/2}{n-k} = \begin{cases} \frac{1}{2^n} \binom{n}{n/2}, & n \text{ is even;} \\ 0, & n \text{ is odd.} \end{cases}$$

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