Find $\sum_{i\in\mathbb{N}}(n-2i)^k\binom{n}{2i+1}$ 
Find $$\sum_{i\in\mathbb{N}}(n-2i)^k\binom{n}{2i+1},$$ where both $n$ and $k$ are natural numbers. 

I know the following identity:
$$
\sum_{i\in\{0\}\cup\mathbb{N}}i(i-1)\cdots(i-p)\binom{n}{2i+1}=(n-p-2)(n-p-3)\cdots(n-2p-2)2^{n-2p-3}.
$$
But I am not sure whether this is helpful.

Suppose we seek to evaluate
$$\sum_{q=0}^n (n-2q)^k {n\choose 2q+1}.$$
We observe that
$$(n-2q)^k = \frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}} \exp((n-2q)z) \; dz.$$
This yields for the sum
$$\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}} 
\sum_{q=0}^n {n\choose 2q+1} \exp((n-2q)z) \; dz
\\ = \frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp((n+1)z)}{z^{k+1}} 
\sum_{q=0}^n {n\choose 2q+1} \exp((-2q-1)z) \; dz$$
which is
$$\frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp((n+1)z)}{z^{k+1}} 
\\ \times 
\left(\sum_{q=0}^n {n\choose q} \exp(-qz)
- \sum_{q=0}^n {n\choose q} (-1)^q \exp(-qz)\right)
\; dz.$$
This yields two pieces, call them $A_1$ and $A_2.$ Piece $A_1$ is
$$\frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp((n+1)z)}{z^{k+1}} 
(1+\exp(-z))^n \; dz
\\ = \frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{k+1}} 
(\exp(z)+1)^n \; dz$$
and piece $A_2$ is
$$\frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp((n+1)z)}{z^{k+1}} 
(1-\exp(-z))^n \; dz
\\ = \frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{k+1}} 
(\exp(z)-1)^n \; dz.$$
Recall the species equation for labelled set partitions:
$$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$
which yields the bivariate generating function of the Stirling numbers
of the second kind
$$\exp(u(\exp(z)-1)).$$
This implies that
$$\sum_{n\ge q} {n\brace q} \frac{z^n}{n!} =
\frac{(\exp(z)-1)^q}{q!}$$
and 
$$\sum_{n\ge q} {n\brace q} \frac{z^{n-1}}{(n-1)!} =
\frac{(\exp(z)-1)^{q-1}}{(q-1)!} \exp(z).$$
Now to evaluate $A_1$ proceed as follows:
$$\frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{k+1}} 
(2+\exp(z)-1)^n \; dz
\\ = \frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{k+1}} 
\sum_{q=0}^n {n\choose q} 2^{n-q} (\exp(z)-1)^q \; dz 
\\ = \sum_{q=0}^n {n\choose q} 2^{n-q} \times 
q!\times \frac{1}{2}\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{k+1}} 
\frac{(\exp(z)-1)^q}{q!} \; dz.$$
Recognizing  the differentiated  Stirling  number generating  function
this becomes
$$\sum_{q=0}^n {n\choose q} 2^{n-q-1} \times 
q! \times {k+1\brace q+1}.$$
Now observe that when $n\gt  k+1$ the Stirling number for $k+1\lt q\le
n$ is zero, so we may replace $n$ by $k+1.$ Similarly, when $n\lt k+1$
the binomial coefficient  for $n\lt q\le k+1$ is zero  so we may again
replace $n$ by $k+1.$ This gives the following result for $A_1:$
$$\sum_{q=0}^{k+1} {n\choose q} 2^{n-q-1} \times 
q! \times {k+1\brace q+1}.$$
Moving on to  $A_2$ we observe that when $k\lt  n$ the contribution is
zero because the  series for $\exp(z)-1$ starts at  $z.$ This integral
is simple and we have
$$\frac{1}{2}\frac{k!\times n!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(z)}{z^{k+1}} 
\frac{(\exp(z)-1)^n}{n!} \; dz.$$
Recognizing the Stirling number this yields
$$\frac{1}{2} \times n! \times  {k+1\brace n+1}.$$
which correctly  represents the fact  that we have a  zero contribution
when $k\lt n.$
This finally yields the closed form formula
$$\sum_{q=0}^{k+1} {n\choose q} 2^{n-q-1} \times 
q! \times {k+1\brace q+1}
- \frac{1}{2} \times n! \times  {k+1\brace n+1}.$$
confirming the previous results.
This MSE link has a computation that is quite similar.
