$\binom{1/2}{n}=\frac{2(-1)^{n-1}}{n4^n}\binom{2n-2}{n-1}$ There is a "direct" way to prove this equality.
$\displaystyle \binom{1/2}{n}=\frac{2(-1)^{n-1}}{n4^n}\binom{2n-2}{n-1}$
I am trying to skip the induction.  Maybe there is a rule or formula that will help me.
Thank you
 A: Just move the numbers around in the formula:
$$\begin{array}{cl}
 \displaystyle\binom{1/2}{n} & \displaystyle = \frac{(\frac{1}{2})(\frac{1}{2}-1)\cdots(\frac{1}{2}-(n-1))}{n!} \\[5pt]
& \displaystyle = \frac{(1)(1-2)(1-4)\cdots(1-2(n-1))}{2^nn!} \\[5pt]
& \displaystyle = (-1)^{n-1}\frac{(1)(3)\cdots(2n-3)}{2^nn!} \\[5pt]
& \displaystyle = (-1)^{n-1}\frac{(1)(2)(3)(4)\cdots(2n-3)(2n-2)}{2^nn!\cdot(2)(4)\cdots(2n-2)} \\[5pt]
& \displaystyle = (-1)^{n-1}\frac{(2n-2)!}{2^nn!\cdot2^n(n-1)!} \\[5pt]
& \displaystyle = (-1)^{n-1}\frac{(2n-2)!}{4^nn\cdot(n-1)!^2} \\[5pt]
& \displaystyle = \frac{(-1)^{n-1}}{4^nn}\binom{2n-2}{n-1}
\end{array} $$
A: another way
Let's rewrite part of the RHS in terms of gamma
$$
\frac{1}{n}\left( \begin{array}{c}
 2n - 2 \\  n - 1 \\ \end{array} \right)
 = \frac{{\Gamma \left( {2n - 1} \right)}}
{{n\Gamma \left( n \right)\Gamma \left( n \right)}}
 = \frac{{\Gamma \left( {2n - 1} \right)}}{{\Gamma \left( {n + 1} \right)\Gamma \left( n \right)}}
$$
To the above we can apply the Gamma Duplication Formula](https://en.wikipedia.org/wiki/Multiplication_theorem#Gamma_function%E2%80%93Legendre_formula)
$$
\Gamma \left( {2\,z} \right)
 = \frac{{2^{\,2\,z - 1} }}{{\sqrt \pi  }}\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right)
 = 2^{\,2\,z - 1} \frac{{\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right)}}{{\Gamma \left( {1/2} \right)}}
$$
wherefrom
$$
\Gamma \left( {2n - 1} \right) = \Gamma \left( {2\left( {n - 1/2} \right)} \right)
 = 2^{\,2n - 2} \frac{{\Gamma \left( {n - 1/2} \right)\Gamma \left( n \right)}}{{\Gamma \left( {1/2} \right)}}
$$
so getting
$$
\begin{array}{l}
 \frac{1}{n}\left( \begin{array}{c}
 2n - 2 \\  n - 1 \\ 
 \end{array} \right) = \frac{{\Gamma \left( {2n - 1} \right)}}{{\Gamma \left( {n + 1} \right)\Gamma \left( n \right)}}
 = 2^{\,2n - 2} \frac{{\Gamma \left( {n - 1/2} \right)}}{{\Gamma \left( {1/2} \right)\Gamma \left( {n + 1} \right)}} =  \\ 
  = 2^{\,2n - 2} \frac{{\Gamma \left( {n - 1/2} \right)}}
{{\left( { - \frac{1}{2}} \right)\Gamma \left( { - 1/2} \right)\Gamma \left( {n + 1} \right)}}
 =  - 2^{\,2n - 1} \left( \begin{array}{c}
 n - 3/2 \\  n \\ 
 \end{array} \right) =  \\ 
  =  - 2^{\,2n - 1} \left( { - 1} \right)^n \left( \begin{array}{c}
 1/2 \\  n \\ 
 \end{array} \right) \\ 
 \end{array}
$$
where:

*

*1st line : application of the duplication formula,

*2nd line: passing from $\Gamma(1/2)$ to $\Gamma(-1/2)$ to write the fraction as a binomial

*3rd line : applying the "upper negation".

