# Binomial Theorem/Abel's Generalisation/Proof 3

## Theorem

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

## Proof

$(1): \quad \paren {x + y}^n = \ds \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}^{n - \epsilon_1 - \cdots - \epsilon_n}$

Setting $z = z_1 = z_2 = \cdots z_n$ we have:

 $\ds$  $\ds \sum x \paren {x + \epsilon_1 z + \cdots + \epsilon_n z}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y - \epsilon_1 z - \cdots - \epsilon_n z}^{n - \epsilon_1 - \cdots - \epsilon_n}$ $\ds$ $=$ $\ds \sum \binom n k x \paren {x + k z}^{k - 1} \paren {y - k z}^{n - k}$ where $\epsilon_1 + \cdots + \epsilon_n = k$ $\ds$ $=$ $\ds \sum \binom n {n - k} x \paren {x + k z}^{k - 1} \paren {y - k z}^{n - k}$ Symmetry Rule for Binomial Coefficients $\ds$ $=$ $\ds \sum \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}$

Hence the result.