Binomial Theorem/Abel's Generalisation/Proof 1

From ProofWiki
Jump to: navigation, search

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 $\left({x + y}\right)$:

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

$\blacksquare$


Sources