What's the coefficent of $x^{20}$ in $(x^3+x^4+x^5+...)^5$? Only hint is needed.

Question

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.

Answer 1

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 $$ \frac{d}{dx}\frac{1}{1-x}=\frac{1}{(1-x)^2}, \quad \frac{d}{dx}\frac{1}{(1-x)^2}=\frac{2}{(1-x)^3}, \quad \frac{d}{dx}\frac{1}{(1-x)^3}=\frac{3!}{(1-x)^4}, \quad \frac{d^n}{dx^n}\frac{1}{1-x}=\frac{(n-1)!}{(1-x)^n} $$

Answer 2

Can be written as

$x^{15}(1-x)^{-5}$

Use binomial expansion for $(1-x)^{-5}$

That is $$x^{15}(1+5x+ \frac{5 \cdot 6}{2!}x^2 + \frac{5 \cdot 6 \cdot 7}{3!}x^3 + \frac{5 \cdot 6 \cdot 7 \cdot 8}{4!}x^4 + \frac{5 \cdot 6 \cdot 7 \cdot 8 \cdot 9}{5!}x^5 + \cdots)$$

Coefficient of $x^{20}$ is

$x^{15}×$ coefficient of $x^5$ .

Which is $$\frac{5 \cdot 6 \cdot 7 \cdot 8 \cdot 9}{5!}$$