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

Theorem

$\ds \paren {x + y}^n = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}$

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

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

$\ds \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-x + k z}^{n - k} = 0$

is to be proved.

So:

$\ds \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-x + k z}^{n - k}$

$=$

$\ds \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-1}^{n - k} \paren {x - k z}^{n - k}$

$\ds $

$=$

$\ds \sum_k \binom n k \paren {-1}^{n - k} x \paren {x - k z}^{n - 1}$

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

where:

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

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

$\ds \sum_k \binom n k \paren {-1}^{n - k} \paren {b_0 + b_1 k + \cdots + b_{n - 1} k^n + b_n 0} = 0$

Hence the result.

$\blacksquare$