# Binomial Theorem/Abel's Generalisation/Proof 1

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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

By admitting $y = \left({x + y}\right) - x$, we have that:

$\left({x + y}\right)^n = \left({x + \left({x + y}\right) - x}\right)^n$

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

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

$\blacksquare$