Coefficient of $x^n$ in the power series form of $\sqrt[3]{(1-2x)}$ What is the coefficient of $x^n$ in the power series form of $\sqrt[3]{(1-2x)}$?
So here's what I've found:
$\sqrt[3]{(1-2x)} = (1-2x)^{1/3}$
Which using Newton's General Binomial Theorem, I get:
$$\sum_{k\geq 0}{{1/3}\choose{k}}(-2x)^k(1)^{{1/3} - k}$$
So the coefficient of $x^n$ is when $k = n$, which is 
$${{1/3}\choose{n}}(-2)^n = (-2)^n\dfrac{1/3(-2/3)(-5/3)\cdots(1/3 - n +1)}{n!}$$
$$= (-1)^n2^n\dfrac{1/3(-2/3)(-5/3)\cdots\left(\frac{1 - 3n +3}{3}\right)}{n!}$$
$$= (-1)^n2^n\dfrac{(-2)(-5)\cdots(-3n +4)}{3^nn!}$$
$$= (-1)^n2^n\dfrac{(-1)^{n-1}(2)(5)\cdots(3n - 4)}{3^nn!}$$
$$= (-1)^{2n-1}2^n\dfrac{(3n - 4)!!!}{3^nn!}$$
$$= -2^n\dfrac{(3n - 4)!!!}{3^nn!}$$ where $n!!! = n(n-3)(n-6)\cdots(n ($mod$ 3))$
My question is, first off, is this correct, or did I make a mistake somewhere. And then, if it is correct, is there something more I can do to simplify this or is this the nicest I can make it look?
 A: It looks perfectly fine to me. As an alternative approach, you may notice that the function $f(x)=\sqrt[3]{1-x}$ fulfills the differential equation
$$ 3 (1-x)\,f'(x) + f(x) = 0.\tag{1} $$
Since $f(x)$ is analytic in a neighbourhood of zero we may also assume $f(x)=\sum_{n\geq 0}a_n x^n$, with $a_0=f(0)=1$, and the previous differential equation turns into a simple recurrence for the coefficients $a_n$:
$$ \frac{a_{n+1}}{a_n} = \frac{3n-1}{3n+3}\tag{2} $$
The conclusion is just the same. This approach is also known as Frobenius' method.
A: Looks OK to me.
Two quibbles.
A minor typo:
the term in the sum
should be $(-1)^k$.
I would write the expansion as
$\begin{array}\\
{{1/3}\choose{n}}(-2)^n 
&= (-2)^n\dfrac{\prod_{k=0}^{n-1}(1/3-k)}{n!}\\
&= (-1)^n2^n\dfrac{\prod_{k=0}^{n-1}(1-3k)/3}{n!}\\
&= (-1)^n2^n\dfrac{\prod_{k=0}^{n-1}(1-3k)}{3^nn!}\\
&= 2^n\dfrac{\prod_{k=0}^{n-1}(3k-1)}{3^nn!}\\
&= -2^n\dfrac{\prod_{k=1}^{n-1}(3k-1)}{3^nn!}\\
&= -2^n\dfrac{(3n-4)!!!}{3^nn!}
\qquad\text{this is optional}\\
\end{array}
$
