I have to represent the function on the left as a power series, and this is the solution to it but I don't know how to calculate this for example when n=1?
Calculation of C(-1/2, n)
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$\begingroup$
combinatorics
functions
power-series
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0$\displaystyle{{-1/2 \choose n} = {n - 3/2 \choose n}\left(-1\right)^{n}}$ – 2017-01-21
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1$\displaystyle{{-1/2 \choose n} = \left(-\,{1 \over 4}\right)^{n} {2n \choose n}}$ – 2017-01-21
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0How do I get that? Did you use some identities? – 2017-01-22
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0I got it in [another post](http://math.stackexchange.com/a/1991085/85343). – 2017-01-23
3 Answers
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The symbol ${\alpha \choose n}$ is defined by :
$${\alpha \choose n}=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}$$
So :
$${{-\frac{1}{2}}\choose0}=1\qquad{{-\frac{1}{2}}\choose1}=-\frac{1}{2}\qquad{{-\frac{1}{2}}\choose2}=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)}{2}=\frac{3}{8}$$
and so on ...
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The formula for $n$ positive integer applies too:
$${n\choose k}=\frac{n(n-1)\cdots(n-k+1)}{k!}$$
regardles whether $n$ is integer or not. It works even for complex $n$. More details here
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By definition
$${-\frac{1}{2} \choose 2}=\frac{-\frac{1}{2}(-\frac{1}{2}-1)}{2!}$$
Notice $2$ terms in numerator.
$${-\frac{1}{2} \choose 3}=\frac{-\frac{1}{2}(-\frac{1}{2}-1)(-\frac{1}{2}-2)}{3!}$$
Notice $3$ terms in numerator.
But also remember ${ -\frac{1}{2} \choose 0}=1$.