<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] $\begin{array}\\
f(x)
&=\dfrac{(1+x)^4}{(1+2x)^3(1-2x)^2}\\
&=\dfrac{(1+x)^4(1-2x)}{(1+2x)^3(1-2x)^3}\\
&=\dfrac{(1+x)^4(1-2x)}{(1-4x^2)^3}\\
&=(1+4x+6x^2+4x^3+x^4)(1-2x)(1-4x^2)^{-3}\\
&=(1+2x-2x^2+...)\sum_{k=0}^{\infty} \binom{k+2}{2})(4x^2)^k
\qquad\text{generalized binomial theorem}\\
&=(1+2x-2x^2+...)(1+3(4x^2)+6(4x^2)^2+...)\\
&=(1+2x-2x^2+...)(1+12x^2+96x^4+...)\\
&=1+2x+10x^2+...\\
\end{array}
$
</s>