Binomial Theorem/Abel's Generalisation/Proof 3

Theorem

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

Proof

From Hurwitz's Generalisation of Binomial Theorem:

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

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.

Sources

1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $51$