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

Theorem

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

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 $s = -\dfrac 1 2$.

This gives:

At this point we can invoke Factors of Binomial Coefficient: Corollary 1 and obtain:

When $k > 0$ we have:

When $k = 0$ we have:

while:

Hence:


 * $(1): \quad \dbinom {1/2} k = \dfrac {2 \left({-1}\right)^{k - 1} } {4^k \left({2 k - 1}\right)} \dbinom {2 k - 1} k - \delta_{k 0}$

where $\delta_{k 0}$ denotes the Kronecker delta.

Then:

We can then make substitutions: