Sum over k of r Choose k by -1^r-k by Polynomial

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 1

From the corollary to Sum over $k$ of $\dbinom r k \dbinom {s + k} n \paren {-1}^{r - k}$:

$\ds \sum_k \binom r k \binom k n \paren {-1}^{r - k} = \delta_{n r}$

where $\delta_{n r}$ denotes the Kronecker delta.

Thus when $n \ne r$:

$\ds \sum_k \binom r k \binom k n \paren {-1}^{r - k} = 0$

and so:

$\ds \sum_k \binom r k \paren {-1}^{r - k} \paren {c_0 \binom k 0 + c_1 \binom k 1 + \cdots + c_m \binom k m} = c_r$

as the only term left standing is the $r$th one.

Choosing the coefficients $c_i$ as appropriate, a polynomial in $k$ can be expressed as a summation of binomial coefficients in the form:

$c_0 \dbinom k 0 + c_1 \dbinom k 1 + \cdots + c_m \dbinom k m$

Thus we can rewrite such a polynomial in $k$ as:

$b_0 + b_1 k + \cdots + b_r k^r$

Since each $c_m \dbinom k m$ is a polynomial of degree $m$, it follows that the only one with a non-zero degree $r$ term is $c_r \dbinom k r$.

The coefficient of $k^r$ in $c_r \dbinom k r$ must be equal to $b_r$, that is:

$b_r = \dfrac {c_r}{r!}$

Hence the result.

$\blacksquare$

Proof 2

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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