Answer
$\dfrac{2}{{4 - 3x}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{3^n}}}{{{2^{2n + 1}}}}{x^n}$, ${\ \ \ }$ converges for $\left| x \right| \lt \dfrac{4}{3}$
Work Step by Step
Write $\dfrac{2}{{4 - 3x}} = \dfrac{2}{{4\left( {1 - \left( {\dfrac{3}{4}x} \right)} \right)}}$.
We substitute $\dfrac{3}{4}x$ for $x$ in Eq. (2) of Section 11.6:
$\dfrac{2}{{4 - 3x}} = \dfrac{2}{{4\left( {1 - \left( {\dfrac{3}{4}x} \right)} \right)}} = \dfrac{1}{2}\left( {\dfrac{1}{{1 - \left( {\dfrac{3}{4}x} \right))}}} \right) = \dfrac{1}{2}\mathop \sum \limits_{n = 0}^\infty {\left( {\dfrac{3}{4}x} \right)^n}$
$\dfrac{2}{{4 - 3x}} = \dfrac{1}{2}\mathop \sum \limits_{n = 0}^\infty \dfrac{{{3^n}}}{{{2^{2n}}}}{x^n} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{3^n}}}{{{2^{2n + 1}}}}{x^n}$
The expansion above is valid for $\left| {\dfrac{3}{4}x} \right| \lt 1$ or $\left| x \right| \lt \dfrac{4}{3}$.
