Could someone help me to prove this identity: $\sum_\limits{k=0}^n \binom{-\frac{1}{2}}{k}\binom{-\frac{1}{2}}{n-k} = (-1)^n$ without using of the infinite series, please? I don't know how to do this without the Vandermonde's identity for real numbers...
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1$\begingroup$ You asked the same question twice. Weren't you satisfied by robjohn's answer? My proposal is to merge both questions into a single one. $\endgroup$ – Jack D'Aurizio Dec 30 '17 at 10:21
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Vandermonde's identity is $$\binom{a+b}n=\sum_{k=0}^n\binom ak\binom b{n-k}.\tag{1}$$ If $a$ and $b$ are non-negative integers (1) has a combinatorial proof. (Count number of $n$-person committees you can make from a pool of $a$ women and $b$ men). The general case can be deduced from this. The difference between the sides in (1) is a polynomial $f(a,b)$ with degree at most $n$ in $a$ and also at most $n$ in $b$. If $a$ is a nonnegative integer then $f(a,0)=f(a,1)=\cdots=f(a,n)=0$ and as $f$ has degree $\le n$ in $b$ then $f(a,b)=0$ for all nonnegative integers $a$ and all $b$. Now for any $b$, $f(0,b)=f(1,b)=\cdots=f(n,b)$. Likewise this implies that $f(a,b)=0$ for all $a$ and $b$, in particular for $a=b=-\frac12$.
answered Dec 30 '17 at 8:11
