I am trying to prove that the following equation is equal to zero.
$$
0=
 \sum\_{j=J+1}^N \Big(j (1-q)+ (j-J) (q N-j) \Big) \cdot q^{j} (1-q)^{N -j} \binom{N}{j} \label{zero1}$$


Where
$J,N \in \mathbb{Z}^+$ and $J<N$ and
$0<q<1$ is a probability.


[Numerical simulations (see link)](https://drive.google.com/file/d/1EZLceMYjPvPQT_WqUe4EON4iOhmKCzkG/view?usp=sharing) suggest that this is true.
Tips and hints (or solutions) are very welcome!



How to prove that these partial binomial sums are zero?
Write $p=1-q$, $qN-j=(N-j)q-jp$, the $j$-th summand is $$(j(1-j+J)p+(j-J)(N-j)q)q^jp^{N-j} {N\choose j}.$$
Expand the brackets and consider it as a homogeneous polynomial in $p$ and $q$ of degree $N+1$. The coefficient of $q^{j+1}p^{N-j}$ equals
$$
(j-J)(N-j){N\choose j}+(j+1)(J-j){N\choose j+1}=0.
$$


