Represent $f(x)=\frac{1}{(1-x^2)^4}$ as a power series

Question

Represent $f(x)=\frac{1}{(1-x^2)^4}$ as a power series

$$\sum\limits_{n=0}^{\infty}x^{2n}=\frac{1}{1-x^2}$$

Second derivative is

$$\left(\frac{1}{1-x^2}\right)^{''}=\frac{1}{(1-x^2)^4}\cdot x(1+8x-2x^2-8x^3+x^4)$$

This gives $$\frac{1}{(1-x^2)^4}=\sum\limits_{n=0}^{\infty}\frac{2n(2n-1)x^{2n-2}}{x(1+8x-2x^2-8x^3+x^4)}$$

I have a proof in combinatorics which involves this series. How to represent this series using binomial coefficients?

Answer 1

Perhaps it could be easier expand $(1-x^2)^{-4}$ using the binomial series.

$\begin{align} (1-x^2)^{-4} &= 1+4x^2 + \frac{(-4)(-4-1)}{2}(-x^2)^2+\frac{(-4)(-4-1)(-4-2)}{3!}(-x^2)^3+ \dots\\ &=1+4x^2+10x^4+20x^6+35x^8+\dots\\ &=\sum_{n=1}^{\infty} \binom{n+2}{3}x^{2n-2}\\ \end{align}$

Answer 2

$$\frac{1}{1-x}=\sum_{n=0}^{\infty }x^n$$ $$\frac{d^3}{dx^3}(\frac{1}{1-x})=\frac{6}{(1-x)^4}=\sum_{n=3}^{\infty }n(n-1)(n-2)x^{n-3}$$ $$\frac{1}{(1-x)^4}=\frac{1}{6}\sum_{n=3}^{\infty }n(n-1)(n-2)x^{n-3}$$ now replace $x$ by $x^2$ $$\frac{1}{(1-x^2)^4}=\frac{1}{6}\sum_{n=3}^{\infty }n(n-1)(n-2)x^{2n-6}=1+4x^2+10x^4+20x^6+35x^8+..$$