<s>[INST] Manipulation of a binomial coefficient In obtaining a formula for the Catalan numbers I have got the expression $-\frac{1}{2}\binom{1/2}{n}(-4)^n$. All my efforts to show that this simplifies to $\frac{1}{n}\binom{2n-2}{n-1}$ have not succeeded. Is there some mistake in the original expression, and if not, how do I simplify it?
Thanks for your time.
 [/INST] First we show that
$$\binom{1/2}{n}={1\over n!}(1/2-1)(1/2-2)...(1/2-(n-1))=$$$$={1\over n!}(-1/2)(-3/2)...(-(2n-3)/2)=$$
$$={1\over n!}{(-1)^{n-1}\over2^n}1\cdot3\cdots(2n-3)\frac{2\cdot4\cdot6...(2n-4)}{2^{n-2}(n-2)!}=$$
$$={1\over n!}{(-1)^{n-1}\over2^{2n-2}}\frac{(2n-3)!}{(n-2)!}\frac{(2n-2)(2n-1)2n}{(n-1)n\cdot2^2\cdot(2n-1)}=$$
$$={1\over n!}{(-1)^{n-1}\over2^{2n}}\frac{(2n)!}{n!(2n-1)}=$$
$$={(-1)^{n-1}\over4^n(2n-1)}\frac{(2n)!}{n!n!}={(-1)^{n-1}\over4^n(2n-1)}\binom{2n}{n}$$ then we get
$$\frac{-1}{2}\binom{1/2}{n}(-4)^n=\frac{-1}{2}{(-1)^{n-1}\over4^n(2n-1)}\binom{2n}{n}(-1)^n4^n=$$
$$={(-1)^{2n}\over2(2n-1)}\binom{2n}{n}={1\over2(2n-1)}{2n(2n-1)\over n(n-1)}\binom{2n-2}{n-2}=$$
$$={1\over n-1}\binom{2n-2}{n-2}={1\over n-1}\binom{2n-2}{n}$$
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