Answer
$f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \left( {{2^{n + 1}} - 1} \right){x^n}$
Work Step by Step
From Table 2, we have
$\dfrac{1}{{1 - x}} = \mathop \sum \limits_{n = 0}^\infty {x^n} = 1 + x + {x^2} + {x^3} + {x^4} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for $\left| x \right| \lt 1$
Substituting $2x$ for $x$ in the series above gives
$\dfrac{1}{{1 - 2x}} = \mathop \sum \limits_{n = 0}^\infty {2^n}{x^n}$, ${\ \ \ }$ converges for $\left| {2x} \right| \lt 1$ or $\left| x \right| \lt \dfrac{1}{2}$
Thus,
$f\left( x \right) = \dfrac{2}{{1 - 2x}} - \dfrac{1}{{1 - x}} = 2\mathop \sum \limits_{n = 0}^\infty {2^n}{x^n} - \mathop \sum \limits_{n = 0}^\infty {x^n}$
The two series together converges for $\left| x \right| \lt \dfrac{1}{2}$.
Therefore,
$f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \left( {{2^{n + 1}} - 1} \right){x^n}$, ${\ \ \ }$ converges for $\left| x \right| \lt \dfrac{1}{2}$.
