Find the closed form of summation of binomial coefficients For positive integers $k$, I have gotten that $$\sum\limits_{i=0}^{k}\frac{(-1)^{k+i-1}}{i+1}\binom{k+i}{i}\binom{k}{i}=0.$$
But for positive integers $m$ with $1\leq m<k$, how can I get the closed form of $$\sum\limits_{i=0}^{m}\frac{(-1)^{k+i-1}}{i+1}\binom{k+i}{i}\binom{k}{i}.$$
 A: Suppose for we seek a closed form of
$$\sum_{q=0}^m \frac{(-1)^{q-1}}{q+1}
{k+q\choose q} {k\choose q}.$$
where $1\le m\lt k.$ This is (Iverson bracket)
$$[z^m] \frac{1}{1-z}
\sum_{q\ge 0} z^q \frac{(-1)^{q-1}}{q+1}
{k+q\choose q} {k\choose q}
\\ = \frac{1}{k+1} [z^m] \frac{1}{1-z}
\sum_{q\ge 0} z^q (-1)^{q-1}
{k+q\choose q} {k+1\choose q+1}
\\ = \frac{1}{k+1} [z^m] \frac{1}{1-z}
[w^k] (1+w)^{k+1}
\sum_{q\ge 0} z^q (-1)^{q-1}
{k+q\choose q} w^q
\\ = \frac{1}{k+1} [z^m] \frac{1}{z-1}
[w^k] (1+w)^{k+1}
\frac{1}{(1+wz)^{k+1}}.$$
The contribution from $w$ is
$$\;\underset{w}{\mathrm{res}}\;
\frac{1}{w^{k+1}} (1+w)^{k+1} \frac{1}{(1+wz)^{k+1}}.$$
We put $w/(1+w) = v$ so that $w=v/(1-v)$ and $dw = 1/(1-v)^2 \; dv$ to
get
$$\;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{k+1}} \frac{1}{(1+zv/(1-v))^{k+1}}
\frac{1}{(1-v)^2}
\\ = \;\underset{v}{\mathrm{res}}\;
\frac{1}{v^{k+1}} \frac{(1-v)^{k-1}}{(1-v(1-z))^{k+1}}.$$
This is
$$\sum_{q=0}^{k-1} (-1)^q {k-1\choose q}
{2k-q\choose k-q} (1-z)^{k-q}.$$
Applying the coefficient extractor in $z$ we find
$$\frac{(-1)^{m-1}}{k+1} \sum_{q=0}^{k-1} (-1)^q {k-1\choose q}
{2k-q\choose k-q} {k-1-q\choose m}.$$
Observe that
$${k-1\choose q} {k-1-q\choose m}
= \frac{(k-1)!}{q! \times m! \times (k-1-q-m)!}
= {k-1\choose m} {k-1-m\choose q}.$$
This will correctly evaluate to zero when $k-1-m\lt q.$ Continuing we
find
$$\frac{(-1)^{m-1}}{k+1} {k-1\choose m}
\sum_{q=0}^{k-1} (-1)^q
{2k-q\choose k-q} {k-1-m\choose q}.$$
Working with the sum we see that we may lower to $q=k-1-m$ due to the
third binomial coefficient and the condition $1\le m\lt k.$ We thus
obtain
$$\sum_{q=0}^{k-1-m} (-1)^q
{2k-q\choose k-q} {k-1-m\choose q}
\\ = [z^k] (1+z)^{2k}
\sum_{q=0}^{k-1-m} (-1)^q
\frac{z^q}{(1+z)^q} {k-1-m\choose q}
\\ =  [z^k] (1+z)^{2k}
\left[1-\frac{z}{1+z}\right]^{k-1-m}
= [z^k] (1+z)^{k+1+m} = {k+1+m\choose k}.$$
Collecting everything we finally have
$$\bbox[5px,border:2px solid #00A000]{
\frac{(-1)^{m+1}}{k+1} {k-1\choose m} {k+1+m\choose k}.}$$
The OP had an extra factor $(-1)^k$ on the sum.
