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Recently I came across a finite sum which appears to be zero for all odd numbers. The sum is defined as follows:

$$\sum_{k=0}^{m}~\binom{n}{k}(n-2k)(-1)^k$$ where $n=2m+1$.

For the first few $m$ this sum always equals zero. I tried to prove this by induction with the inductive step from $m=k$ to $m=k+1$ but I did not got this far with this approach. So I am asking for a proof of this formula with an explanation.


I have found more of these sums and I would be interested in a more general result. It seems to me like that in general the exponent of the term $n-2k$ is irrelevant as long as it is an odd number. So that

$$\sum_{k=0}^{m}~\binom{n}{k}(n-2k)^3(-1)^k$$ $$\sum_{k=0}^{m}~\binom{n}{k}(n-2k)^5(-1)^k$$ $$...$$

all equal zero. I guess thats a fact but I have no clue how to derivate this.


I have to add that it is not needed for the number to be odd or even - all these sums should work out for both. So also the following should equal zero.

$$\sum_{k=0}^{m}~\binom{n}{k}(n-2k)^2(-1)^k$$ $$\sum_{k=0}^{m}~\binom{n}{k}(n-2k)^4(-1)^k$$ $$...$$

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We have that for $n=2m+1>1$, then $$\begin{align} \sum_{k=0}^{m}~\binom{n}{k}(n-2k)(-1)^k&=\sum_{k=0}^{m}~\binom{n}{k}(n-k)(-1)^k-\sum_{k=0}^{m}~\binom{n}{k}k(-1)^k\\ &=\sum_{k'=n-m}^{n}~\binom{n}{n-k'}k'(-1)^{n-k'}-\sum_{k=0}^{m}~\binom{n}{k}k(-1)^k\\ &=-\sum_{k'=m+1}^{n}~\binom{n}{k'}k'(-1)^{k'}-\sum_{k=0}^{m}~\binom{n}{k}k(-1)^k\\ &=-\sum_{k=0}^{n}~\binom{n}{k}k(-1)^{k}=n\sum_{k=1}^{n}~\binom{n-1}{k-1}(-1)^{k-1}\\ &=n(1+(-1))^{n-1}=0\end{align}$$ where at the second step, we rewrite the first sum with respect to the new index $k':=n-k$.

P.S. If $d$ is odd then again by letting $k'=n-k$, $$\sum_{k=0}^{m}\binom{n}{k}(n-2k)^d(-1)^k=\sum_{k'=m+1}^{n}\binom{n}{n-k'}(n-2(n-k'))^d(-1)^{n-k'}\\=\sum_{k=m+1}^{n}\binom{n}{k'}(n-2k')^d(-1)^{k'}$$ which implies that $$\sum_{k=0}^{m}\binom{n}{k}(n-2k)^d(-1)^k=\frac{1}{2}\sum_{k=0}^{n}\binom{n}{k}(n-2k)^d(-1)^k.$$ This sum is NOT always zero. For example, when $n=d$, by Tepper's identity, we have that $$\sum_{k=0}^{n}\binom{n}{k}(n-2k)^n(-1)^k=2^n\cdot n!.$$

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  • $\begingroup$ I am not quite sure how about the index change from $k=0$ to $k=n-m$ and then further to $k=m+1$. Could you explain this in detail? $\endgroup$ – mrtaurho Jul 19 '18 at 14:48
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    $\begingroup$ If we replace $k$ with $n-k'$ then the new index $k'=n-k$ goes from $n-0$ to $n-m=m+1$ because $k$ goes from $0$ to $m$. $\endgroup$ – Robert Z Jul 19 '18 at 14:54
  • $\begingroup$ Okay, now I realized that this is clear to see. But I am still confused about the substitution of $k$ with $n-k$. I see why do you did it there, but how does $k=0$ become $k=n-m$ and how does $m$ become $n$. My first assumption would be an index shift up by $m$. But how does that effect on the rest of the sum? $\endgroup$ – mrtaurho Jul 19 '18 at 17:17
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    $\begingroup$ Let the new index be $k'=n-k$. Then for $k=0$, $k'=n$, for $k=1$, $k'=n-1$,..., for $k=m$, $k'=n-m=2m+1-m=m+1$. I edited my answer. Is it better now? $\endgroup$ – Robert Z Jul 19 '18 at 18:16
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With the expansion $$(1-x)^n=\sum_{k=0}^{n}{n\choose k}(-x)^{k}$$ then derivate respect to $x$, and separate the sum into equal number terms: \begin{align} -n(1-x)^{n-1} &= \sum_{k=0}^{n}{n\choose k}(-k)(-x)^{k-1} \\ &= \sum_{k=0}^{2m+1}{n\choose k}(-k)(-x)^{k-1} \\ &= \sum_{k=0}^{m}{n\choose k}(-k)(-x)^{k-1} + \sum_{k=m+1}^{2m+1}{n\choose k}(-k)(-x)^{k-1} \\ &= \sum_{k=0}^{m}{n\choose k}(-k)(-x)^{k-1} + \sum_{k=0}^{m}{n\choose k+m+1}-(k+m+1)(-x)^{k+m} ~~~ \text{now let} ~~ k\to m-k \\ &= \sum_{k=0}^{m}{n\choose k}(-k)(-x)^{k-1} + \sum_{k=0}^{m}{n\choose 2m+1-k}-(2m+1-k)(-x)^{2m-k} \\ &= \sum_{k=0}^{m}{n\choose k}(-k)(-1)^{k-1}x^{k-1} + \sum_{k=0}^{m}{n\choose k}-(2m+1-k)(-1)^{k}x^{2m-k} \\ \end{align} set $x=1$ then $$\color{blue}{0}= \sum_{k=0}^{m}{n\choose k}(k)(-1)^{k} + \sum_{k=0}^{m}{n\choose k}-(2m+1-k)(-1)^{k}= \color{blue}{\sum_{k=0}^{m}{n\choose k}(2k-n)(-1)^{k}}$$

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