Calculation of C(-1/2, n)

Question

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?

Answer 1

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

Answer 2

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 ${ -\frac{1}{2} \choose 0}=1$.

Answer 3

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 ...

Answer 4

$${-\frac12\choose n}=\frac{(\tfrac12-1)(\tfrac12-2)...(\tfrac12-n)}{n!}=\frac{(-1)^n(2n-1)!!}{n!2^n}=\frac{(-1)^n(2n)!}{n!2^n n!2^n}$$ So, $$\color{blue}{{-\frac12\choose n}=\left(-\frac14\right)^n{2n\choose n}}$$ And also $$\color{red}{{\frac12\choose n}=\frac1{1-2n}{-\frac12\choose n}}$$