Finite Binomial Sum I am pretty sure the following sum is zero:
$\sum_{i=0}^{\lfloor \frac{n+1}{2} \rfloor} \binom{n+2}{2 i +1}(4i-n)$
and have proven it to be zero for even $n$ (simply reverse the sum with variable substitution $i=n/2-j$ and then use the property that $\binom{n}{k}=\binom{n}{n-k}$ and show that the sum is equal to its negative.)
However I cannot figure out how to do the case where $n$ is odd.  Mathematica seems to say that:
$\sum_{i=0}^{\lfloor \frac{n+1}{2} \rfloor} \binom{n+2}{2 i +1}(4i-n)=\frac{-2}{1+n} \binom{n+2}{1+2(1+\lfloor \frac{n+1}{2} \rfloor)}(1+\lfloor \frac{n+1}{2} \rfloor)(3+2\lfloor \frac{n+1}{2} \rfloor)$
but I have no idea what kind of identities or properties it is using to get that result.
 A: 
Setting $n=2N-1$ we obtain
  \begin{align*}
\color{blue}{\sum_{i=0}^N}&\color{blue}{\binom{2N+1}{2i+1}(4i-2N+1)}\\
&=\sum_{i=0}^N\binom{2N+1}{2i+1}(4i+2)-(2N+1)\sum_{i=0}^N\binom{2N+1}{2i+1}\tag{1}\\
&=2(2N+1)\sum_{i=0}^N\binom{2N}{2i}-(2N+1)\sum_{i=0}\left[\binom{2N}{2i+1}+\binom{2N}{2i}\right]\tag{2}\\
&=(2N+1)\left[\sum_{i=0}^N\binom{2N}{2i}-\sum_{i=0}^{N-1}\binom{2N}{2i+1}\right]\tag{3}\\
&=(2N+1)\sum_{i=0}^{2N}\binom{2N}{i}(-1)^i\tag{4}\\
&=(2N+1)(1+(-1))^{2N}\tag{5}\\
&\,\,\color{blue}{=0}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we split the sum conveniently.

*In (2) we apply in the left sum the binomial identity $\binom{p+1}{q+1}=\frac{p+1}{q+1}\binom{p}{q}$ and in the right sum the binoimial identity $\binom{p+1}{q+1}=\binom{p}{q+1}+\binom{p}{q}$.

*In (3) we collect equal summands.

*In (4) we observe we can merge the left sum containing the even parts and the right sum which contains the odd parts of $(-1)^i\binom{2N}{i}$. Note that now the upper limit of the sum is $2N$.

*In (5) we apply the binomial theorem.
