Sum over k of r Choose k by -1^r-k by Polynomial/Proof 2

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Theorem

Let $r \in \Z_{\ge 0}$.

Then:

$\ds \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$

where:

$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a polynomial in $k$ of degree $r$.


Proof

From Summation of Powers over Product of Differences:

$\ds \sum_{j \mathop = 1}^n \begin {pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end {pmatrix} = \begin {cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end {cases}$


Now we have:

\(\ds \frac {\paren {-1}^k \dbinom n k} {n!}\) \(=\) \(\ds \frac {\paren {-1}^k} {k! \paren {n - k}!}\)
\(\ds \) \(=\) \(\ds \frac {\paren {-1}^k} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {k - j} }\)


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