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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.

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  • $\begingroup$ Try putting $2m+1$ instead of $n$ and $m+1$ instead of the floor function. Then ask Mathematica to FullSimplify after it gives that mess. It will return 0. $\endgroup$ – almagest Feb 23 '18 at 10:46
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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.

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  • $\begingroup$ Thanks! I have actually been trying to transform it into a binomial partial sum, but this way is way more elegant. $\endgroup$ – Ben Feb 24 '18 at 22:39
  • $\begingroup$ @Ben: You're welcome. $\endgroup$ – Markus Scheuer Feb 24 '18 at 22:40

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