Answer
$f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \left[ {{{\left( { - 1} \right)}^{n + 1}}{{\left( {\dfrac{2}{3}} \right)}^{n + 1}} + {{\left( { - 1} \right)}^n}} \right]{\left( {x - 2} \right)^n}$,
valid for $\left| {x - 2} \right| \lt 1$.
Work Step by Step
Recall the identity in Exercise 69:
$f\left( x \right) = \dfrac{2}{{1 - 2x}} - \dfrac{1}{{1 - x}}$
We can rewrite this as
$f\left( x \right) = \dfrac{2}{{ - 3 - 2\left( {x - 2} \right)}} - \dfrac{1}{{ - 1 - \left( {x - 2} \right)}}$
$f\left( x \right) = \dfrac{2}{{ - 3\left( {1 + \dfrac{{x - 2}}{{3/2}}} \right)}} + \dfrac{1}{{\left( {1 + \left( {x - 2} \right)} \right)}}$
From Table 2, we have
(1) ${\ \ \ \ \ }$ $\dfrac{1}{{1 + x}} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{x^n} = 1 - x + {x^2} - {x^3} + {x^4} - \cdot\cdot\cdot$,
converges for $\left| x \right| \lt 1$.
Substituting $\dfrac{{x - 2}}{{3/2}}$ for $x$ in the expansion (1) gives
$\dfrac{1}{{1 + \dfrac{{x - 2}}{{3/2}}}} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{\left( {\dfrac{{x - 2}}{{3/2}}} \right)^n}$, ${\ \ \ }$ converges for $\left| {\dfrac{{x - 2}}{{3/2}}} \right| \lt 1$ or $\left| {x - 2} \right| \lt \dfrac{3}{2}$
So,
$\dfrac{1}{{1 + \dfrac{{x - 2}}{{3/2}}}} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{\left( {\dfrac{2}{3}} \right)^n}{\left( {x - 2} \right)^n}$
Substituting $\left( {x - 2} \right)$ for $x$ in the expansion (1) gives
$\dfrac{1}{{\left( {1 + \left( {x - 2} \right)} \right)}} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{\left( {x - 2} \right)^n}$, ${\ \ \ }$ converges for $\left| {x - 2} \right| \lt 1$
Therefore,
$f\left( x \right) = \dfrac{2}{{ - 3\left( {1 + \dfrac{{x - 2}}{{3/2}}} \right)}} + \dfrac{1}{{\left( {1 + \left( {x - 2} \right)} \right)}}$
$f\left( x \right) = - \dfrac{2}{3}\mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{\left( {\dfrac{2}{3}} \right)^n}{\left( {x - 2} \right)^n} + \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{\left( {x - 2} \right)^n}$
$f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^{n + 1}}{\left( {\dfrac{2}{3}} \right)^{n + 1}}{\left( {x - 2} \right)^n} + \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{\left( {x - 2} \right)^n}$
The two series together is valid for $\left| {x - 2} \right| \lt 1$.
Thus, $f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \left[ {{{\left( { - 1} \right)}^{n + 1}}{{\left( {\dfrac{2}{3}} \right)}^{n + 1}} + {{\left( { - 1} \right)}^n}} \right]{\left( {x - 2} \right)^n}$.
