Theorem: Product of r Choose k with r Minus Half Choose k/Formulation 1

Let $k \in \Z$, $r \in \R$.
:$\dbinom r k \dbinom {r - \frac 1 2} k = \dfrac {\dbinom {2 r} k \dbinom {2 r - k} k} {4^k}$
where $\dbinom r k$ denotes a binomial coefficient.

Proof:
From Binomial Coefficient expressed using Beta Function:
:$(1): \quad \dbinom r k \dbinom {r - \frac 1 2} k = \dfrac 1 {\paren {r + 1} \map \Beta {k + 1, r - k + 1} \paren {r + \frac 1 2} \map \Beta {k + 1, r - k + \frac 1 2} }$
Then:
{{begin-eqn}}
{{eqn | l = \dbinom r {k + 1} \dbinom {r - \frac 1 2} {k + 1}
      | r = \dfrac 1 {\paren {r + 1} \map \Beta {k + 2, r - k} \paren {r + \frac 1 2} \map \Beta {k + 2, r - k - \frac 1 2} }
      | c = 
}}
{{eqn | r = \dfrac 1 {\paren {r + 1} \frac {k + 1} {r + 1} \map \Beta {k + 1, r - k} \paren {r + \frac 1 2} \frac {k + 1} {r + \frac 1 2} \map \Beta {k + 1, r - k - \frac 1 2} }
      | c = Beta Function of $x$ with $y+1$ by $\dfrac {x+y} y$
}}
{{eqn | r = \dfrac 1 {\paren {r + 1} \frac {k + 1} {r + 1} \frac {r + 1} {r - k} \map \Beta {k + 1, r - k + 1} \paren {r + \frac 1 2} \frac {k + 1} {r + \frac 1 2} \frac {r + \frac 1 2} {r - k - \frac 1 2} \map \Beta {k + 1, r - k + \frac 1 2} }
      | c = Beta Function of $x$ with $y+1$ by $\dfrac {x+y} y$
}}
{{eqn | r = \dfrac {\paren {r - k} \paren {r - k  - \frac 1 2} } {\paren {k + 1}^2} \times \frac 1 {\paren {r + 1} \map \Beta {k + 1, r - k + 1} \paren {r + \frac 1 2} \map \Beta {k + 1, r - k + \frac 1 2} }
      | c = simplifying
}}
{{eqn | n = 2
      | r = \dfrac {\paren {r - k} \paren {r - k  - \frac 1 2} } {\paren {k + 1}^2} \dbinom r k \dbinom {r - \frac 1 2} k
      | c = from $(1)$
}}
{{end-eqn}}
Again from Binomial Coefficient expressed using Beta Function:
:$(3): \quad \dbinom {2 r} k \dbinom {2 r - k} k = \dfrac 1 {\paren {2 r + 1} \map \Beta {k + 1, 2 r - k + 1} \paren {2 r - k + 1} \map \Beta {k + 1, 2 r - 2 k + 1} }$
Then:
{{begin-eqn}}
{{eqn | l = \dbinom {2 r} {k + 1} \dbinom {2 r - k - 1} {k + 1}
      | r = \dfrac 1 {\paren {2 r + 1} \map \Beta {k + 2, 2 r - k} \paren {2 r - k} \map \Beta {k + 2, 2 r - 2 k - 1} }
      | c = 
}}
{{eqn | r = \dfrac 1 {\paren {2 r + 1} \frac {k + 1} {2 r + 1} \map \Beta {k + 1, 2 r - k} \paren {2 r - k} \frac {k + 1} {2 r - k} \map \Beta {k + 1, 2 r - 2 k - 1} }
      | c = Beta Function of $x$ with $y+1$ by $\dfrac {x+y} y$
}}
{{eqn | r = \dfrac 1 {\paren {k + 1}^2 \map \Beta {k + 1, 2 r - k} \map \Beta {k + 1, 2 r - 2 k - 1} }
      | c = simplification
}}
{{eqn | r = \dfrac 1 {\paren {k + 1}^2 \frac {2 r + 1} {2 r - k} \map \Beta {k + 1, 2 r - k + 1} \frac {2 r - k} {2 r - 2 k - 1} \map \Beta {k + 1, 2 r - 2 k} }
      | c = Beta Function of $x$ with $y+1$ by $\dfrac {x+y} y$
}}
{{eqn | r = \dfrac {\paren {2 r - 2 k - 1} } {\paren {k + 1}^2 \paren {2 r + 1} \map \Beta {k + 1, 2 r - k + 1} \map \Beta {k + 1, 2 r - 2 k} }
      | c = simplifying
}}
{{eqn | r = \dfrac {\paren {2 r - 2 k - 1} } {\paren {k + 1}^2 \paren {2 r + 1} \map \Beta {k + 1, 2 r - k + 1} \frac {2 r - k + 1} {2 r - 2 k} \map \Beta {k + 1, 2 r - 2 k + 1} }
      | c = Beta Function of $x$ with $y+1$ by $\dfrac {x+y} y$
}}
{{eqn | r = \dfrac {\paren {2 r - 2 k - 1} \paren {2 r - 2 k} } {\paren {k + 1}^2} \times \frac 1 {\paren {2 r + 1} \map \Beta {k + 1, 2 r - k + 1} \paren {2 r - k + 1} \map \Beta {k + 1, 2 r - 2 k + 1} }
      | c = simplifying
}}
{{eqn | r = \dfrac {\paren {2 r - 2 k} \paren {2 r - 2 k - 1} } {\paren {k + 1}^2} \dbinom {2 r} k \dbinom {2 r - k} k
      | c = from $(3)$
}}
{{eqn | r = \dfrac {4 \paren {r - k} \paren {r - k - \frac 1 2} } {\paren {k + 1}^2} \dbinom {2 r} k \dbinom {2 r - k} k
      | c = 
}}
{{end-eqn}}
{{finish}}