consider the following example: $$ \left(1+x\right)^{\frac{1}{2}}=\sum_{k=0}^{∞ } {{\frac{1}{2}}\choose{k}} \cdot x^k={{\frac{1}{2}}\choose{0}}x^0+{{\frac{1}{2}}\choose{1}}x^1+{{\frac{1}{2}}\choose{2}}x^2+...= $$$$\frac{\left(\frac{1}{2}\right)!}{\left(\frac{1}{2}\right)!(0)!}x^0+\frac{\left(\frac{1}{2}\right)!}{\left(-\frac{1}{2}\right)!(1)!}x^1+\frac{\left(\frac{1}{2}\right)!}{\left(-\frac{3}{2}\right)!(2)!}x^2+...$$ as it's clear here we have two negative factorial,but why we have this error?
now look at this one which is just a substitution:
let $a=1/2$, then we have: $\left(1+x\right)^{a}=\sum_{k=0}^{∞ } {{a}\choose{k}} \cdot x^k={{a}\choose{0}}x^0+{{a}\choose{1}}x^1+{{a}\choose{2}}x^2+...=\frac{(a)!}{(a)!(0)!}x^0+\frac{(a)!}{(a-1)!(1)!}x^1+\frac{(a)!}{(a-2)!(2)!}x^2+...=1+ax+\frac{a\left(a-1\right)}{2}x^2+...$$
substitute back $a=1/2$ finally $=1+\frac{1}{2}x-\frac{1}{8}x^2+...$
this answer true but actually it depends on the first one , so the question is why the first one does not give the true answer?
Extending Dr Zafar Ahmed DSc'a answer, since $\binom{p}{k} =\dfrac{\prod_{j=0}^{k-1}(p-j)}{k!} $,
$\begin{array}\\ \binom{\frac12}{k} &=\dfrac{\prod_{j=0}^{k-1}(\frac12-j)}{k!}\\ &=\dfrac{\prod_{j=0}^{k-1}(1-2j)/2}{k!}\\ &=\dfrac{\prod_{j=0}^{k-1}(1-2j)}{2^kk!}\\ &=(-1)^k\dfrac{\prod_{j=0}^{k-1}(2j-1)}{2^kk!}\\ &=(-1)^{k+1}\dfrac{\prod_{j=1}^{k-1}(2j-1)}{2^kk!}\\ &=(-1)^{k+1}\dfrac{\prod_{j=1}^{k-1}(2j-1)\prod_{j=1}^{k-1}(2j)}{2^kk!\prod_{j=1}^{k-1}(2j)}\\ &=(-1)^{k+1}\dfrac{\prod_{j=1}^{2k-2}j}{2^{2k-1}k!(k-1)!}\\ &=(-1)^{k+1}\dfrac{(2k-2)!}{2^{2k-1}k!(k-1)!}\\ &=(-1)^{k+1}\dfrac{(2k)!}{2^{2k-1}k!(k-1)!(2k-1)(2k)}\\ &=(-1)^{k+1}\dfrac{(2k)!}{2^{2k}k!^2(2k-1)}\\ &=\dfrac{(-1)^{k+1}\binom{2k}{k}}{2^{2k}(2k-1)}\\ \end{array} $
$k=0 \to 1,\\ k=1 \to \dfrac{\binom{2}{1}}{2^2\cdot 1} =\dfrac{1}{2},\\ k=2 \to \dfrac{\binom{4}{2}}{2^4\cdot 3} =-\dfrac{6}{48} =-\dfrac{1}{8},\\ k=3 \to \dfrac{\binom{6}{3}}{2^6\cdot 5} =\dfrac{20}{320} =\dfrac{1}{16}\\ $
Note: none of this is original.