Analytic Function Problem Differentiation

Question

If $$F(z)=\sum_{k = 0}^{\infty}{\alpha \choose k}z^k$$ We have to show that $$(1+z) F'(z)=\alpha F(z)$$

well to solve this I will first compute $F'$ and multiply it by $(1+z)$. But I am getting stuck in computing $F'(z)$.

Answer 1

$$ \binom{\alpha}{k} = \frac{\alpha(\alpha-1) \dotsm (\alpha-k+1)}{k!} = \frac{\alpha}{k} \frac{(\alpha-1) \dotsm (\alpha-k+1)}{(k-1)!} = \frac{\alpha}{k} \binom{\alpha-1}{k-1}, $$ so $$ F'(z) = \sum_{k=1}^{\infty} \binom{\alpha}{k} kx^{k-1} = \alpha\sum_{k=1}^{\infty} \binom{\alpha-1}{k-1} x^{k-1} = \alpha\sum_{n=0}^{\infty} \binom{\alpha-1}{n} x^{n}. $$ Now multiply by $1+z$ and use Pascal's identity $$ \binom{\alpha-1}{n}+\binom{\alpha-1}{n-1} = \binom{\alpha}{n} $$ to get to the result.