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I know a proof of the first and of the third of the following equalities. I am looking for a proof of the second equality.

$\sum\limits_{i=0}^{n} \;(-1)^{n - i} \cdot \binom{n}i = 0$

$\sum\limits_{i=0}^{n} \;(-1)^{n - i} \cdot \binom{n}i \cdot i^k = 0$ for $k$ with $0 < k < n$

$\sum\limits_{i=0}^{n} \;(-1)^{n - i} \cdot \binom{n}i \cdot i^n = n!$

darij grinberg
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Loic
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  • Hint: there are two ways of doing this kind of thing. (1) think of a clever combinatorial argument; (2) use induction. If you cannot quickly do (1), try (2). Try adding $S_n$ to $S_{n+1}$ term by term What do you get? – almagest Sep 25 '19 at 17:04
  • Theorem 0.2 on UMN Spring 2018 Math 4707 homework set #3 says that $\sum\limits_{k=0}^c \left(-1\right)^k \dbinom{c}{k} \left(b-k\right)^a = 0$ for any $a \in \mathbb{N}$, $b \in \mathbb{Q}$ and $c \in \mathbb{N}$ satisfying $c > a$. If you reindex the sum by substituting $i$ for $b-k$, then this becomes $\sum\limits_{i=0}^c \left(-1\right)^{c-i} \dbinom{c}{i} \left(b-c+i\right)^a = 0$ (since $\dbinom{c}{c-i} = \dbinom{c}{i}$). Now set $c = n$, $b = n$ and $a = k$, and you get your second identity. – darij grinberg Oct 02 '19 at 16:43

1 Answers1

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It is convenient to use the coefficient of operator to denote the coefficient of $z^k$ of a series. This way we can write for instance \begin{align*} [z^k](1+z)^n=\binom{n}{k} \qquad \text{and}\qquad k![z^k]e^{qz}=k![z^k]\sum_{j=0}^\infty \frac{(qz)^j}{j!}=q^k\tag{1} \end{align*}

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

Comment:

  • In (2) we use the coefficient of operator according to (1).

  • In (3) we apply the binomial theorem.

  • In (4) we see the expansion gives powers of $z$ starting with $z^n$.

Markus Scheuer
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    Nice approach. The development above actually holds for any integral $k$, proving all three equalities at once. And more: for instance, for $k = n + 1$, the sum is thus $n * (n + 1)! / 2$. – Loic Sep 25 '19 at 21:36