<s>[INST] Represent $f(x)=\frac{1}{(1-x^2)^4}$ as a power series 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?
 [/INST] $$\frac{1}{1-x}=\sum_{n=0}^{\infty }x^n$$
$$\frac{d^3}{dx^3}(\frac{1}{1-x})=\frac{6}{(1-x)^4}=\sum_{n=3}^{\infty }n(n-1)(n-2)x^{n-3}$$
$$\frac{1}{(1-x)^4}=\frac{1}{6}\sum_{n=3}^{\infty }n(n-1)(n-2)x^{n-3}$$
now replace $x$ by $x^2$
$$\frac{1}{(1-x^2)^4}=\frac{1}{6}\sum_{n=3}^{\infty }n(n-1)(n-2)x^{2n-6}=1+4x^2+10x^4+20x^6+35x^8+..$$
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