Using the binomial theorem I'd like some help with proving the next equation: 
$$\sqrt{1+x}=\sum_{0}^{\infty }\frac{(-1)^{n-1}}{2^{2n-1}\cdot n}\binom{2n-2}{n-1}\cdot x^{n}$$
 A: Let's rewrite the expression in terms of the Gamma function
$$
\sqrt {1 + x}  = \sum\limits_{0 \leqslant n} {\frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2^{\,2n - 1} n}}\left( \begin{gathered}
  2(n - 1) \\ 
  n - 1 \\ 
\end{gathered}  \right)x^{\,n} }  = \sum\limits_{0 \leqslant n} {\frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2^{\,2n - 1} n}}\frac{{\Gamma (2n - 1)}}
{{\Gamma (n)^{\,2} }}x^{\,n} } 
$$
and then consider the Duplication formula
$$
\Gamma \left( {2z} \right) = \frac{{2^{\,2\,z - 1} }}
{{\sqrt \pi  }}\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right) = \frac{{2^{\,2\,z - 1} }}
{{\Gamma \left( {1/2} \right)}}\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right)
$$
so that the second term in the summand will become
$$
\frac{{\Gamma (2n - 1)}}
{{\Gamma (n)^{\,2} }} = \frac{{\Gamma \left( {2\left( {n - 1/2} \right)} \right)}}
{{\Gamma (n)^{\,2} }} = \frac{{2^{\,2\,n - 2} }}
{{\Gamma \left( {1/2} \right)}}\frac{{\Gamma \left( {n - 1/2} \right)}}
{{\Gamma \left( n \right)}}
$$
Therefore the whole coefficient of $x^n$ reduces to:
$$
\begin{gathered}
  \frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2^{\,2n - 1} n}}\left( \begin{gathered}
  2(n - 1) \\ 
  n - 1 \\ 
\end{gathered}  \right) = \frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2^{\,2n - 1} n}}\frac{{\Gamma (2n - 1)}}
{{\Gamma (n)^{\,2} }} =  \hfill \\
   = \frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2^{\,2n - 1} n}}\frac{{2^{\,2\,n - 2} }}
{{\Gamma \left( {1/2} \right)}}\frac{{\Gamma \left( {n - 1/2} \right)}}
{{\Gamma \left( n \right)}} =  \hfill \\
   = \frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2n}}\frac{{\Gamma \left( {n - 1/2} \right)}}
{{\Gamma \left( {1/2} \right)\Gamma \left( n \right)}} = \frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2n}}\left( \begin{gathered}
  n - 1 - 1/2 \\ 
  n - 1 \\ 
\end{gathered}  \right) =  \hfill \\
   = \frac{1}
{{2n}}\left( \begin{gathered}
   - 1/2 \\ 
  n - 1 \\ 
\end{gathered}  \right) = \frac{{\frac{1}
{2}\Gamma (1/2)}}
{{n\;\Gamma (n)\;\Gamma (1 + 1/2 - n)}} = \frac{{\Gamma (1 + 1/2)}}
{{\Gamma (1 + n)\;\Gamma (1 + 1/2 - n)}} =  \hfill \\
   = \left( \begin{gathered}
  1/2 \\ 
  n \\ 
\end{gathered}  \right) \hfill \\ 
\end{gathered} 
$$
Note that the Duplication formula stems from splitting the product into even and odd components
as follows
$$
\begin{gathered}
  \prod\limits_{0\, \leqslant \,k\; \leqslant \,n} {\left( {x + k} \right)}  = \prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {n/2} \right\rfloor } {\left( {x + 2j} \right)} \;\prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor } {\left( {x + 2j + 1} \right)}  =  \hfill \\
   = 2^{\,\left\lfloor {n/2} \right\rfloor  + 1} \prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {n/2} \right\rfloor } {\left( {x/2 + j} \right)} \;\;2^{\,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor  + 1} \prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor } {\left( {x/2 + 1/2 + j} \right)}  =  \hfill \\
   = \prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {n/2} \right\rfloor } {\left( {x + 2j} \right)\prod\limits_{0\, \leqslant \,j\; \leqslant \,\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor } {\left( {x - 1 + 2\left( {j + 1} \right)} \right)} }  =  \hfill \\
   = \quad \; \cdots  \hfill \\ 
\end{gathered} 
$$
with all the various "manipulations" you can do on the terms and on the multiplication bounds.
A: For $|x|<1$ we have (Newton)$$(1+x)^{1/2}=1+x(1/2)/1!+x^2(1/2)(1/2-1)/2!+x^3(1/2)(1/2-1)(1/2-2)/3!+...$$  The coefficients $A_n$ of $x^n$ in this series satisfy the conditions $A_0=1$ and $A_{n+1}=A_n(1/2-n)/(n+1).$
Defining $\binom {-2}{-1}=1$ (so the term with $n=0$ in your series makes sense) you can   confirm that the coefficients in your series satisfy these conditions also so it is exactly Newton's series. 
