Summation from k to m of 2k-1 Choose k by 2n-2k Choose n-k by -1 over 2k-1

Theorem

 * $\displaystyle \sum_{k \mathop = 0}^m \binom {2 k - 1} k \binom {2 n - 2 k} {n - k} \dfrac {-1} {2 k - 1} = \dfrac {n - m} {2 n} \dbinom {2 m} m \dbinom {2 n - 2 m} {n - m} + \dfrac 1 2 \dbinom {2 n} n$

Proof
From Summation from k to m of r Choose k by s Choose n-k by nr-(r+s)k:


 * $\displaystyle \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$

Set $r - \dfrac 1 2$ and $s = -\dfrac 1 2$.


 * $\displaystyle \sum_{k \mathop = 0}^m \dbinom {1/2} k \dbinom {-1/2} {n - k} \paren {\frac n 2 - \paren {\frac 1 2 - \frac 1 2} k} = \paren {m + 1} \paren {n - m} \dbinom {1/2} {m + 1} \dbinom {-1/2} {n - m}$

Take the :

Now the :

Thus we have: