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?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
\mrm{f}\pars{x} & \equiv 
{\pars{1 + x}^{4} \over \pars{1 + 2x}^{3}\pars{1 - 2x}^{2}}
\,\,\,\stackrel{\mrm{as}\ x\ \to\ 0}{\sim}\,\,\,
{1 + 4x + 6x^{2} \over
\pars{1 + 6x + 12x^{2}}\pars{1 - 4x + 4x^{2}}}
\\[5mm] & \stackrel{\mrm{as}\ x\ \to\ 0}{\sim}\,\,\,
{1 + 4x + 6x^{2} \over
1 - 4x + 4x^{2} + 6x - 24x^{2} + 12x^{2}}
\\[5mm] & = 
\pars{1 + 4x + 6x^{2}}\pars{1 + 2x - 8x^{2}}^{-1}
\\[5mm] & \stackrel{\mrm{as}\ x\ \to\ 0}{\sim}\,\,\,
\pars{1 + 4x + 6x^{2}}\bracks{1 - \pars{2x - 8x^{2}} + \pars{2x - 8x^{2}}^{2}}
\\[5mm] & \stackrel{\mrm{as}\ x\ \to\ 0}{\sim}\,\,\,
\pars{1 + 4x + 6x^{2}}\pars{1 - 2x + 8x^{2} + 4x^{2}}
\\[5mm] & =
\pars{1 + 4x + 6x^{2}}\pars{1 - 2x + 12x^{2}}
\\[5mm] & \stackrel{\mrm{as}\ x\ \to\ 0}{\sim}\,\,\,
1 - 2x + 12x^{2} + 4x - 8x^{2} + 6x^{2} = 1 + 2x + 10x^{2}
\end{align}

$$
\bbx{\mrm{f}\pars{x} \equiv 
{\pars{1 + x}^{4} \over \pars{1 + 2x}^{3}\pars{1 - 2x}^{2}}
\,\,\,\stackrel{\mrm{as}\ x\ \to\ 0}{\sim}\,\,\,
\bbox[10px,#ffd]{1 + 2x + 10x^{2}}}
$$
A: $\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}
$
A: Just for the fun of it !
Starting from @marty cohen's answer and continuing the process
$$f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}=\sum_{n=0}^\infty  2^{n-8} \left[27 (6 n+7)+(-1)^n (2 n^3+13n+67)\right]\,x^n$$
