A problem related to binomial theorem for any index

Question

If $|x|<1$, then in the expansion of $(1+2x+3x^2+4x^3+\cdots)^{1/2}$, the coefficient of $x^n$ is I tried to do it with the AGP approach but wasn't able to solve it completely because of the exponential power

Answer 1

Note that $1+x+x^2+x^3+...=\frac{1}{1-x}$. Then $1+2x+3x^2+...=\left(\frac{1}{1-x}\right)'=\frac{1}{(1-x)^2}$. Then $(1+2x+3x^2+...)^{1/2}=\frac{1}{1-x}=1+x+x^2+x^3+...$

Answer 2

Like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $

we can prove for $|x|<1,$ $$S=\sum_{r=0}^\infty rx^{r-1}=(1-x)^{-2}$$

Alternatively $$S=\sum_{r=0}\dfrac{d(x^r)}{dx}=\dfrac{d\left(\sum_{r=0}^\infty x^r\right)}{dx}$$ which is for $|x|<1,$ $$=\dfrac{d(1-x)^{-1}}{dx}=?$$