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 derive 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.

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


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 $$\sum_{k=0}^{n}\binom{n}{k}(n-2k)^n(-1)^k=2^n\cdot n!.$$

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $$\color{blue}{0}= \sum_{k=0}^{m}{n\choose k}(k)(-1)^{k} + \sum_{k=0}^{m}{n\choose k}-(2m+1-k)(-1)^{k}= \color{blue}{\sum_{k=0}^{m}{n\choose k}(2k-n)(-1)^{k}}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.