If $f(x) = \sum_{n=0}^{\infty}{\alpha\choose n}x^n$, show (without assuming the results of the binomial theorem) that $$(1+x)f'(x)=\alpha f(x)$$ for $|x|<1$
I've already shown that the sum converges, but I have no idea how to manipulate it to get that $\alpha$ in front of the sum.
We have:
$$\begin{align}&f(x)=\sum_{n=0}^\infty\binom\alpha n x^n\\{}\\ &f'(x)=\sum_{n=1}^\infty\binom\alpha n nx^{n-1}\end{align}$$
Perhaps the trick you're looking for is
$$n\binom\alpha n=\alpha\binom{\alpha-1}{n-1}$$
Added on request: $$\alpha f(x)=\sum_{n=0}^\infty \alpha\binom\alpha nx^n=\alpha\binom\alpha0+\alpha\binom\alpha1x+\alpha\binom\alpha2x^2+\ldots$$
OTOH:
$$(1+x)f'(x)=\sum_{n=0}^\infty n\binom \alpha nx^{n-1}+\sum_{n=1}^\infty n\binom\alpha nx^n=$$
$$=\sum_{n=1}^\infty n\binom\alpha nx^{n-1}(1+x)=\sum_{n=1}^\infty\alpha\binom{\alpha-1}{n-1}x^{n-1}(1+x)=$$
$$=\alpha\binom{\alpha-1}0(1+x)+\alpha\binom{\alpha-1}1x(1+x)+\alpha\binom{\alpha-1}2x^2(1+x)+\ldots$$
Well, now just compare coefficients of corresponding powers of $\;x\;$ in both expressions: they are equal.