<s>[INST] Find the third expression of Taylor's for $f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}$ around $x=0$ 
Find the third expression of Taylor's for $f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}$ around $x=0$

My try:Let $y=x+1$. Then: $$f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}=\frac{1}{(2y-1)(4-\frac{1}{y^2})^2}=-\frac{1}{2}\cdot\frac{1}{1-2y}\cdot\frac{1}{(1-\frac{1}{4y^2})^2}=$$ $$=-\frac{1}{2}\sum(2y)^n\sum(\frac{1}{4y^2})^n$$That is why I have $$f^{(3)}=3!\frac{-1}{2}\cdot2\cdot\frac{1}{4}=\frac{1}{4}$$Why is it not correct solution?
 [/INST] Just for the fun of it !
Starting from @marty cohen's answer and continuing the process
$$f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}=\sum_{n=0}^\infty  2^{n-8} \left[27 (6 n+7)+(-1)^n (2 n^3+13n+67)\right]\,x^n$$
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