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

$\blacksquare$

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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 $51$