<s>[INST] What's the coefficent of $x^{20}$ in $(x^3+x^4+x^5+...)^5$? Only hint is needed. I know about Binomial/Multinomial expansion but I got stuck on this series; it doesn't look like anything I've solved before. I already searched for any hint/formula and couldn't find one, any help is appreciated.
 [/INST] Observe that
$$
(x^3+x^4+\cdots+x^n+\cdots)^5=x^{15}(1+x+x^2+\cdots)^5=\frac{x^{15}}{(1-x)^5}\\ =\frac{x^{15}}{3!}\frac{d^4}{dx^4}\frac{1}{1-x}= \frac{x^{15}}{4!}
\left(1+x+x^2+\cdots\right)^{(4)}\\=\frac{x^{15}}{4!}
\left(\frac{4!}{0!}+\frac{5!}{1!}x+\frac{6!}{2!}x^2+\frac{7!}{3!}x^3+\frac{8!}{4!}x^4+\frac{9!}{5!}x^5+\cdots\right)
$$
Hence coefficient of $x^{20}$ is 
$$
c_{20}=\frac{9!}{4!5!}
$$
Note that
$$
\frac{d}{dx}\frac{1}{1-x}=\frac{1}{(1-x)^2}, \quad
\frac{d}{dx}\frac{1}{(1-x)^2}=\frac{2}{(1-x)^3}, \quad
\frac{d}{dx}\frac{1}{(1-x)^3}=\frac{3!}{(1-x)^4}, \quad 
\frac{d^n}{dx^n}\frac{1}{1-x}=\frac{(n-1)!}{(1-x)^n}
$$
</s>