# Binomial Theorem/Abel's Generalisation/Proof 1

## 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

By admitting $y = \paren {x + y} - x$, we have that:

$\paren {x + y}^n = \paren {x + \paren {x + y} - x}^n$

Expanding the right hand side in powers of $\paren {x + y}$:

 $\ds$  $\ds \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}$ $\ds$ $=$ $\ds \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {x + \paren {x + y} + k z}^{n - k}$ $\ds$ $=$ $\ds \sum_k \binom n k x \paren {x - k z}^{k - 1} \sum_j \paren {x + y}^j \paren {-x + k z}^{n - k - j} \binom {n - k} j$ $\ds$ $=$ $\ds \sum_j \binom n j \paren {x + y})^j \sum_k \binom {n - j} {n - j - k} x \paren {x - k z}^{k - 1} \paren {-x + k z}^{n - k - j}$ $\ds$ $=$ $\ds \sum_{j \mathop \le n} \binom n j \paren {x + y}^j 0^{n - j}$ Abel's Generalisation of Binomial Theorem: Special Case $x + y = 0$ $\ds$ $=$ $\ds \paren {x + y}^n$ Binomial Theorem

$\blacksquare$