Product of r Choose k with r Minus Half Choose k/Formulation 1/Proof 1

Theorem

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

First we establish the following:

\(\ds \paren {r - \frac 1 2}^{\underline k}\)

\(=\)

\(\ds \paren {r - \frac 1 2} \paren {r - \frac 3 2} \paren {r - \frac 5 2} \dotsm \paren {r - \frac 1 2 - k + 1}\)

Definition of Falling Factorial

\(\ds \)

\(=\)

\(\ds \dfrac {2^k \paren {r - \frac 1 2} \paren {r - \frac 3 2} \paren {r - \frac 5 2} \cdots \paren {r - \frac 1 2 - k + 1} } {2^k}\)

\(\ds \)

\(=\)

\(\ds \dfrac {\paren {2 r - 1} \paren {2 r - 3} \paren {2 r - 5} \dotsm \paren {2 r - 2 k + 1} } {2^k}\)

\(\ds \)

\(=\)

\(\ds \dfrac {2 r \paren {2 r - 1} \paren {2 r - 2} \paren {2 r - 3} \dotsm \paren {2 r - 2 k + 2} \paren {2 r - 2 k + 1} } {2^k \paren {2 r \paren {2 r - 2} \paren {2 r - 4} \dotsm \paren {2 r - 2 k + 2} } }\)

\(\text {(1)}: \quad\)

\(\ds \)

\(=\)

\(\ds \dfrac {\paren {2 r}^{\underline{2 k} } } {2^k \times 2^k r^{\underline k} }\)

Definition of Falling Factorial

Then:

\(\ds \dbinom r k \dbinom {r - \frac 1 2} k\)

\(=\)

\(\ds \dfrac {r^{\underline k} } {k!} \dfrac {\paren {r - \frac 1 2}^{\underline k} } {k!}\)

Definition of Binomial Coefficient

\(\ds \)

\(=\)

\(\ds \dfrac {r^{\underline k} } {k!} \dfrac {\paren {2 r}^{\underline {2 k} } } {k! \, 2^{2 k} r^{\underline k} }\)

from $(1)$

\(\ds \)

\(=\)

\(\ds \dfrac {\paren {2 r}^{\underline {2 k} } } {\paren {k!}^2 \, 4^k}\)

simplifying

\(\ds \)

\(=\)

\(\ds \dfrac {\paren {2 r}^{\underline k} \paren {2 r - k}^{\underline k} } {\paren {k!}^2 \, 4^k}\)

Falling Factorial of Sum of Integers

\(\ds \)

\(=\)

\(\ds \dfrac 1 {4^k} \dfrac {\paren {2 r}^{\underline k} } {k!} \dfrac {\paren {2 r - k}^{\underline k} } {k!}\)

separating out

\(\ds \)

\(=\)

\(\ds \dfrac 1 {4^k} \dbinom {2 r} k \dbinom {2 r - k} k\)

Definition of Binomial Coefficient