Binomial Theorem/Abel's Generalisation/Proof 1
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 $\left({x + y}\right)$:
| \(\displaystyle \) | \(\) | \(\displaystyle \sum_k \binom n k x \left({x - k z}\right)^{k - 1} \left({y + k z}\right)^{n - k}\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sum_k \binom n k x \left({x - k z}\right)^{k - 1} \left({x + \left({x + y}\right) + k z}\right)^{n - k}\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sum_k \binom n k x \left({x - k z}\right)^{k - 1} \sum_j \left({x + y}\right)^j \left({-x + k z}\right)^{n - k - j} \binom {n - k} j\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sum_j \binom n j \left({x + y}\right)^j \sum_k \binom {n - j} {n - j - k} x \left({x - k z}\right)^{k - 1} \left({-x + k z}\right)^{n - k - j}\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{j \mathop \le n} \binom n j \left({x + y}\right)^j 0^{n - j}\) | $\quad$ Abel's Generalisation of Binomial Theorem: Special Case $x + y = 0$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left({x + y}\right)^n\) | $\quad$ Binomial Theorem | $\quad$ |
$\blacksquare$
In particular: to the results used above.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.
To discuss this in more detail, feel free to use the talk page.
Once you have done so, you can remove this instance of {{MissingLinks}} from the code.
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 $51$
