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I tried to calculate

$ F(m)=\sum\limits_{k=0}^n(-1)^{n-k}C_n^k k^m,$

where $m=0,1,\cdots,m$, $C_n^k$ is the binomial coefficient $\dfrac{n!}{k!(n-k)!}$.

It's obvious when $k$ is $1,2,3$ by Binomial theorem and differentiation.

e.g. $F(2)=\sum\limits_{k=0}^n(-1)^{n-k}C_n^k k^2 x^{k-1}|_{x=1}=(x\cdot \sum\limits_{k=0}^n(-1)^{n-k} C_n^k k x^{k-1})'|_{k=1}=(x\cdot(\sum\limits_{k=0}^n (-1)^{n-k}C_n^k x^k)')'|_{k=1}=(x\cdot((x-1)^n)')'|_{x=1}=(nx(x-1)^{n-1})'|_{x=1}=[n(x-1)^{n-1}+n(n-1)x(x-1)^{n-2}]|_{x=1}=0.$

And repeat this method then I got

$F(m)=0, 0\leqslant m\leqslant n-1.$

I guess that $F(n)=n!$ because when calculating the general $F(m),m<n$ the last term equals to $\dfrac{n!}{(n-m)!}x^{m-1}(x-1)^{n-m}$. So when $m=n$, the other terms have the factor of $(x-1)$, let $x=1$ then they all equal to 0 except the last one. Let $m=n$, it follows that $F(n)=n!$.

But I'm not sure if my proof is rigorous. And I can not calculate the general $F(m)$ directly.

nonuser
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闫嘉琦
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1 Answers1

1

We can show the identity with the help of generating functions. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance \begin{align*} [z^m]e^{kz}=[z^m]\left(1+kz+\frac{(kz)^2}{2!}+\cdots\right)=\frac{k^m}{m!}\tag{1} \end{align*}

We obtain for integers $0\leq m\leq n$ \begin{align*} \color{blue}{\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}k^m}&=\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}m![z^m]e^{kz}\tag{2}\\ &=m![z^m]\sum_{k=0}^n\binom{n}{k}\left(e^z\right)^k(-1)^{n-k}\tag{3}\\ &=m![z^m]\left(e^z-1\right)^n\tag{4}\\ &=m![z^m]\left(z+\frac{z^2}{2!}+\cdots\right)^n\tag{5}\\ &\color{blue}{=\begin{cases} n!\qquad& n=m\\ 0\qquad& 0\leq m<n \end{cases}} \end{align*} and the claim follows.

Comment:

  • In (2) we use the coefficient of operator as shown in (1).

  • In (3) we do a small rearrangement.

  • In (4) we apply the binomial theorem.

  • In (5) we see the smallest power of $z$ is $n$.

Markus Scheuer
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