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 |
$\blacksquare$
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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 $51$