User Did, did this step in his answer to my previous question:
$$\sum_{k=0}^n{n\choose k}(zp)^kq^{n-k}=(q+pz)^n.$$
How is it done? Is it simply an identity, or something more?
2026-03-28 10:03:32.1774692212
How is this step completed?
70 Views Asked by user142198 https://math.techqa.club/user/user142198/detail At
1
There are 1 best solutions below
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Unsure why @Pedro didn't just post an answer, I guess I will (figured it'd be best to close this question).
Check out the Binomial Theorem.
The TL;DR version of this is that given some $n \in \mathbb{N}, a,b \in \mathbb{C}$ we have $$ (a+b)^n = \sum_{i=0}^n \binom{n}{i}a^ib^{n-i} $$ where $$ \binom{n}{i} = \frac{n!}{(n-i)!i!} $$ and $k!$ is defined recursively as: $$ 0! = 1, (k+1)! = (k+1)k! $$