Krdytkyu

I'm trying to proof that
$$sum_{k=l}^{2n} binom{2n+k}{2k} frac{(2k-1)!!}{(k-l)!} (-1)^k = begin{cases} 0 quad {rm if} , , l , ,{rm odd} \ frac{(-1)^{n-l/2}(2n+l)!}{4^n left(n-frac{l}{2}right)! left(n+frac{l}{2}right)!} quad {rm if} , , l , ,{rm even} end{cases}$$
where $()!$ is factorial and $()!!$ double-factorial.
Any idea?

edit: Before down-voting it would be more helpful to state what information is missing, but that is probably too much to ask...

We seek to evaluate

$$sum_{k=q}^{2n} {2n+kchoose 2k}
frac{(2k-1)!}{(k-1)! times 2^{k-1}} frac{1}{(k-q)!} (-1)^k
\ = sum_{k=q}^{2n} {2n+kchoose 2k}
frac{(2k)!}{k! times 2^{k}} frac{1}{(k-q)!} (-1)^k
.$$

This is

$$q! sum_{k=q}^{2n} {2n+kchoose 2k}
frac{(2k)!}{k! times k! times 2^{k}}
frac{k!}{q!times(k-q)!} frac{(-1)^k}{2^k}
\ = q! sum_{k=q}^{2n} {2n+kchoose 2k}
{2kchoose k} {kchoose q}
frac{(-1)^k}{2^k}.$$

Observe that

$${2n+kchoose 2k} {2kchoose k}
= frac{(2n+k)!}{(2n-k)! times k! times k!}
= {2n+kchoose 2n} {2nchoose k}$$

and furthermore

$${2nchoose k} {kchoose q}
= frac{(2n)!}{(2n-k)! times q! times (k-q)!}
= {2nchoose q} {2n-qchoose k-q}.$$

We get for the sum

$${2nchoose q} q!
sum_{k=q}^{2n} {2n+kchoose 2n} {2n-qchoose k-q}
frac{(-1)^k}{2^k}
\ = {2nchoose q} q! frac{(-1)^q}{2^q}
sum_{k=0}^{2n-q} {2n+q+kchoose 2n} {2n-qchoose k}
frac{(-1)^k}{2^k}
.$$

This becomes

$${2nchoose q} q! frac{(-1)^q}{2^q}
sum_{k=0}^{2n-q} {2n+q+kchoose 2n} [z^{2n-q-k}] (1+z)^{2n-q}
frac{(-1)^k}{2^k}
\ = {2nchoose q} q! frac{(-1)^q}{2^q}
[z^{2n-q}] (1+z)^{2n-q}
sum_{k=0}^{2n-q} {2n+q+kchoose 2n}
frac{(-1)^k}{2^k} z^k.$$

Now we may extend $k$ beyond $2n-q$ because of the coefficient
extractor $[z^{2n-q}]$ (no contribution) and get

$${2nchoose q} q! frac{(-1)^q}{2^q}
[z^{2n-q}] (1+z)^{2n-q}
sum_{kge 0} {2n+q+kchoose 2n}
frac{(-1)^k}{2^k} z^k
\ = {2nchoose q} q! frac{(-1)^q}{2^q}
[z^{2n-q}] (1+z)^{2n-q} [w^{2n}] (1+w)^{2n+q}
sum_{kge 0} (1+w)^k
frac{(-1)^k}{2^k} z^k
\ = {2nchoose q} q! frac{(-1)^q}{2^q}
[z^{2n-q}] (1+z)^{2n-q} [w^{2n}] (1+w)^{2n+q}
frac{1}{1+z(1+w)/2}.$$

Re-write this as

$${2nchoose q} q! frac{(-1)^q}{2^q}
[w^{2n}] (1+w)^{2n+q}
mathrm{Res}_{z=0} frac{1}{z^{2n-q+1}} (1+z)^{2n-q}
frac{1}{1+z(1+w)/2}.$$

Working with the residue we apply the substitution $z/(1+z) = v$
or $z=v/(1-v)$ to get

$$mathrm{Res}_{v=0} frac{1}{v^{2n-q}}
frac{1-v}{v} frac{1}{1+(v/(1-v))(1+w)/2} frac{1}{(1-v)^2}
\ = mathrm{Res}_{v=0} frac{1}{v^{2n-q+1}}
frac{1}{1-v+v(1+w)/2}
\ = mathrm{Res}_{v=0} frac{1}{v^{2n-q+1}}
frac{1}{1-v(1-w)/2}
= frac{1}{2^{2n-q}} (1-w)^{2n-q}.$$

Substitute into the remaining coefficient extractor to get

$${2nchoose q} q! frac{(-1)^q}{2^q}
[w^{2n}] (1+w)^{2n+q} frac{1}{2^{2n-q}} (1-w)^{2n-q}
\ = {2nchoose q} q! frac{(-1)^q}{2^{2n}}
sum_{p=0}^{2n-q} (-1)^p {2n-qchoose p} {2n+qchoose 2n-p}.$$

Now

$${2nchoose q} {2n-qchoose p}
= frac{(2n)!}{q!times p! times (2n-q-p)!}
= {2nchoose p} {2n-pchoose q}$$

and

$${2n-pchoose q} {2n+qchoose 2n-p}
= frac{(2n+q)!}{q! times (2n-p-q)! times (p+q)!}
= {2n+qchoose q} {2nchoose p+q}.$$

This yields

$${2n+qchoose q} q! frac{(-1)^q}{2^{2n}}
sum_{p=0}^{2n-q} (-1)^p {2nchoose p} {2nchoose p+q}
\ = {2n+qchoose q} q! frac{(-1)^q}{2^{2n}}
sum_{p=0}^{2n-q} (-1)^p {2nchoose p}
[z^{2n-p-q}] (1+z)^{2n}
\ = {2n+qchoose q} q! frac{(-1)^q}{2^{2n}}
[z^{2n-q}] (1+z)^{2n}
sum_{p=0}^{2n-q} (-1)^p {2nchoose p} z^p.$$

Now we may extend $p$ beyond $2n-q$ because of the coeffcient
extractor $[z^{2n-q}]$ in front. We find

$${2n+qchoose q} q! frac{(-1)^q}{2^{2n}}
[z^{2n-q}] (1+z)^{2n}
sum_{pge 0} (-1)^p {2nchoose p} z^p
\ = {2n+qchoose q} q! frac{(-1)^q}{2^{2n}}
[z^{2n-q}] (1+z)^{2n} (1-z)^{2n}
\ = {2n+qchoose q} q! frac{(-1)^q}{2^{2n}}
[z^{2n-q}] (1-z^2)^{2n}.$$

Concluding we immediately obtain zero when $q$ is odd, and otherwise
we find

$${2n+qchoose q} q! frac{(-1)^q}{2^{2n}}
[z^{2(n-q/2)}] (1-z^2)^{2n}
\ = {2n+qchoose q} q! frac{(-1)^q}{2^{2n}}
[z^{n-q/2}] (1-z)^{2n}.$$

This is

$${2n+qchoose q} q! frac{(-1)^q}{2^{2n}}
(-1)^{n-q/2} {2nchoose n-q/2}$$

or alternatively
$$bbox[5px,border:2px solid #00A000]{
frac{(-1)^{n+q/2}}{2^{2n}}
frac{(2n+q)!}{(n-q/2)! times (n+q/2)!}.}$$