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