Alternating sum of binomial coefficients weighted with the some constant power of their index

Question

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 $0 < k < n$

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

Answer 1

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 $z^n$.