Binomial Theorem/Abel's Generalisation/Proof 2

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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

From this formula:

$(1): \quad \ds \sum_{k \mathop \in \Z} \binom n k x \paren {x + k z}^{k - 1} y \paren {y + \paren {n - k} z}^{n - k - 1} = \paren {x + y} \paren {x + y + n z}^{n - 1}$

The given formula:

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

can then be transformed into $(1)$ by



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