Binomial Theorem/Abel's Generalisation/x+y = 0

Theorem
Consider Abel's Generalisation of Binomial Theorem:

This holds in the special case where $x + y = 0$.

Proof
As $x + y = 0$, we can substitute $y = -x$, and so:


 * $\displaystyle \sum_k \binom n k x \left({x - k z}\right)^{k - 1} \left({-x + k z}\right)^{n - k} = 0$

is to be proved.

So:

Then Sum over $k$ of $\dbinom r k$ by $\left({-1}\right)^{r-k}$ by Polynomial can be applied:


 * $\displaystyle \sum_k \binom r k \left({-1}\right)^{r - k} P_r \left({k}\right) = r! \, b_r$

where:
 * $P_r \left({k}\right) = b_0 + b_1 k + \cdots + b_r k^r$ is a polynomial in $k$ of degree $r$.

As the coefficient of $k^n = 0$, we have:


 * $\displaystyle \sum_k \binom n k \left({-1}\right)^{n - k} \left({b_0 + b_1 k + \cdots + b_{n - 1} k^n + b_n 0}\right) = 0$

Hence the result.