Given the problem:
What is the coefficient of $x^{2005}$ in the generating function $G(x) = \frac{1}{(1-x)^2(1+x)^2}$?
The solution posted starts with:
Let $\frac{1}{(1-x)^2(1+x)^2} = \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{C}{1+x} + \frac{D}{(1+x)^2}$. Upon simplification, the right hand side becomes:
$$\frac{(C-A)x^3 + (B+D-A-C)x^2 + (A+2B-C-2D)x + (A+B+C+D)}{(1-x)^2(1+x)^2}$$
What step-by-step solution did he follow in order to come up with this kind of ordered simplification? I'm sure he used extended binomial theorem to do this but can anyone give me a proper explanation how he applied that to this problem?
We have \begin{align} &\frac A{1-x} + \frac B{(1-x)^2} + \frac C{1+x} + \frac D{(1+x)^2}\\ &=\frac{A(1-x)(1+x)^2 + B(1+x)^2 + C(1-x)^2(1+x) + D(1-x)^2}{(1-x)^2(1+x)^2}\\ &=\frac{A(1+x-x^2-x^3) + B(1+2x+x^2) + C(x^3-x^2-x+1) + D(1-2x+x^2)}{(1-x)^2(1+x)^2}\\ &=\frac{(-A+C)x^3 + (-A+B-C+D)x^2 + (A+2B-C-2D)x + (A+B+C+D)}{(1-x)^2(1+x)^2}. \end{align} and hence $$(-A+C)x^3 + (-A+B-C+D)x^2 + (A+2B-C-2D)x + (A+B+C+D) = 1$$ From this we get the system of linear equations \begin{align} -A+C &= 0\\ -A+B-C+D&=0\\ A+2B-C-2D&=0\\ A+B+C+D&=1 \end{align} which yields $$ A=B=C=D = \frac14.$$ Thus the generating function is \begin{align} &\frac14\sum_{n=0}^\infty x^n + \frac14\sum_{n=0}^\infty (n+1)x^n + \frac14\sum_{n=0}^\infty (-1)^n x^n + \sum_{n=0}^\infty(-1)^n (n+1)x^n\\ &= \sum_{n=0}^\infty \frac14 \left(1 + (-1)^n\right)(n+2)x^{2n}\\ &= \sum_{n=0}^\infty (n+1)x^{2n} \end{align} and clearly the coefficient of $x^{2005}$ is zero.
A much easier way to solve this problem, though, is to recognize that $$(1-x)^2(1+x)^2 = ((1-x)(1+x))^2 = (1-x^2)^2. $$