Prove $\sum_{j=0}^n \left(-\frac{1}{2}\right)^j \binom{n}{j}\binom{n+j}{j}\binom{j}{k} = 0$ when $n+k$ is odd An integral led me to a power series with these coefficients:
$$a_k = \sum_{j=k}^n \left(-\frac{1}{2}\right)^j \binom{n}{j}\binom{n+j}{j}\binom{j}{k}$$
I strongly suspect that the series should have $a_k = 0$ when $n+k$ is odd, and I've verified it for $k,n\leq 10$. I'm looking for a direct proof of this. Does anyone have a suggestion?

Suppose we seek to verify that
$$a_{n,k} = \sum_{j=k}^n \left(-\frac{1}{2}\right)^j
{n\choose j} {n+j\choose j} {j\choose k}$$
is zero when $n+k$ is odd. We have
$${n+j\choose j} {j\choose k}
= \frac{(n+j)!}{n! k! (j-k)!}
= {n+k\choose k} {n+j\choose n+k}$$
and obtain for the sum
$${n+k\choose k} \sum_{j=k}^n \left(-\frac{1}{2}\right)^j
{n\choose j} {n+j\choose n+k}$$
Introduce 
$${n+j\choose n+k}  = {n+j\choose j-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{j-k+1}} (1+z)^{n+j} \; dz$$
Note that this vanishes when $j\lt k$  so we may lower $j$ to start at
zero, getting for the inner sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} z^{k-1} (1+z)^{n} 
\sum_{j=0}^n \left(-\frac{1}{2}\right)^j
{n\choose j} \frac{(1+z)^j}{z^j}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} z^{k-1} (1+z)^{n} 
\left(1-\frac{1+z}{2z}\right)^n
\; dz
\\ = \frac{1}{2^n} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} (1+z)^{n} 
(z-1)^n\; dz
\\ = \frac{1}{2^n} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} (z^2-1)^n
\; dz.$$
We thus have the closed form
$$\frac{1}{2^n} {n+k\choose k} [z^{n-k}] (z^2-1)^n
\\ = \begin{cases}
\frac{1}{2^n} {n+k\choose k} (-1)^{(n+k)/2} {n\choose (n-k)/2}
\quad\text{if}\quad n-k\quad\text{is even}
\\ 0\quad\text{otherwise.}
\end{cases}.$$
We  see having reached  the result  that we  did not  make use  of the
differential in  the integral which  means the above also  works using
formal power series only.
