Binomial Theorem/Abel's Generalisation/Proof 1

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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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