Answer
$f\left( x \right) = \dfrac{1}{{{{\left( {1 - 2x} \right)}^2}}} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - 2}\\
n
\end{array}} \right){\left( { - 1} \right)^n}\dfrac{{{2^n}}}{{{5^{n + 2}}}}{\left( {x + 2} \right)^n}$
Work Step by Step
Write
$\dfrac{1}{{{{\left( {1 - 2x} \right)}^2}}} = \dfrac{1}{{{{\left( {1 - 2\left( {x + 2} \right) + 4} \right)}^2}}} = \dfrac{1}{{{{\left( {5 - 2\left( {x + 2} \right)} \right)}^2}}} = \dfrac{1}{{{5^2}{{\left( {1 - \dfrac{2}{5}\left( {x + 2} \right)} \right)}^2}}} = \dfrac{1}{{{5^2}}}{\left( {1 - \dfrac{2}{5}\left( {x + 2} \right)} \right)^{ - 2}}$
So, $\dfrac{1}{{{{\left( {1 - 2x} \right)}^2}}} = \dfrac{1}{{{5^2}}}{\left( {1 - \dfrac{2}{5}\left( {x + 2} \right)} \right)^{ - 2}}$.
From Table 2 of Section 11.7, we know that
${\left( {1 + x} \right)^a} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){x^n}$,
converges for $\left| x \right| \lt 1$
Setting $a=-2$ and substituting $ - \dfrac{2}{5}\left( {x + 2} \right)$ for $x$ in the series above, we obtain
$\dfrac{1}{{{{\left( {1 - 2x} \right)}^2}}} = \dfrac{1}{{{5^2}}}{\left( {1 - \dfrac{2}{5}\left( {x + 2} \right)} \right)^{ - 2}} = \dfrac{1}{{{5^2}}}\mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - 2}\\
n
\end{array}} \right){\left( { - \dfrac{2}{5}\left( {x + 2} \right)} \right)^n}$
$ = \dfrac{1}{{{5^2}}}\mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - 2}\\
n
\end{array}} \right){\left( { - 1} \right)^n}\dfrac{{{2^n}}}{{{5^n}}}{\left( {x + 2} \right)^n}$
$ = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - 2}\\
n
\end{array}} \right){\left( { - 1} \right)^n}\dfrac{{{2^n}}}{{{5^{n + 2}}}}{\left( {x + 2} \right)^n}$
Thus, the Taylor series centered at $c=-2$ is
$f\left( x \right) = \dfrac{1}{{{{\left( {1 - 2x} \right)}^2}}} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - 2}\\
n
\end{array}} \right){\left( { - 1} \right)^n}\dfrac{{{2^n}}}{{{5^{n + 2}}}}{\left( {x + 2} \right)^n}$
