Prove that:
$$(\alpha+\beta)^n = \sum_{k=0}^{\infty} \binom{n}{k} (\alpha + \beta - 1)^{k} \quad \text{when } |\alpha+\beta-1|<1$$
I got the latter expression from Wolfram Alpha in Series Representation and tried asking ChatGPT to provide a proof (blundered and wrong proof):
To prove this equality, we can use the binomial theorem. The binomial theorem states that for any real numbers α, β, and a positive integer n:
$$(\alpha + \beta)^n = \sum_{k=0}^{n} \binom{n}{k} \alpha^{n-k} \beta^{k}$$
Now, if we let $\alpha' = \alpha$ and $\beta' = \beta - 1$, we can rewrite the binomial theorem as:
$$(\alpha + (\beta - 1))^n = \sum_{k=0}^{n} \binom{n}{k} \alpha^{n-k} (\beta - 1)^{k}$$
Simplifying this expression gives us:
$$(\alpha + \beta - 1)^n = \sum_{k=0}^{n} \binom{n}{k} \alpha^{n-k} (\beta - 1)^{k}$$
This matches the second series you provided. Therefore, the two series are equivalent: $$\sum_{k=0}^{\infty} \binom{n}{k} \alpha^{n-k} \beta^{k} = \sum_{k=0}^{\infty} \binom{n}{k} (\alpha + \beta - 1)^{k}$$
As for now, I have tried proving basic identities using binomial theorem:
$$ (\alpha+\beta)^n = \sum_{k=0}^{\infty} \alpha^{n-k} \beta^k \binom{n}{k} $$
$$ (\alpha+\beta)^2 = \sum_{k=0}^{\infty} \binom{2}{k} \alpha^{2-k} \beta^k $$
$$ = \binom{2}{0} \alpha^2 + \binom{2}{1} \alpha \beta + \binom{2}{2} \beta^2 $$
$$ = \alpha^2 + 2 \alpha \beta + \beta^2 $$
and then this expression:
$$ (\alpha+\beta)^n = \sum_{k=0}^{\infty} (\alpha+\beta-1)^k \binom{n}{k} $$
$$ (\alpha+\beta)^2 = \sum_{k=0}^{\infty} \binom{2}{k} (\alpha+\beta-1)^k $$
$$ = \binom{2}{2} (\alpha+\beta-1)^0 + \binom{2}{1} (\alpha+\beta-1) + \binom{2}{2} (\alpha+\beta-1)^2 $$
$$ = 1 + 2 (\alpha+\beta-1) + (\alpha^2 + \beta^2 + 1 + 2(\alpha \beta - \alpha \beta - \beta)) $$
$$ = \alpha^2 + \beta^2 + 2 \alpha \beta $$