Answer
$\frac{1}{{1 + 2x}}$
Work Step by Step
Write $\mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{2^n}{x^n} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{\left( {2x} \right)^n}$.
Let $t=2x$. So,
$\mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{2^n}{x^n} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{t^n}$
From Table 2, we see that
$\dfrac{1}{{1 + t}} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{t^n}$, ${\ \ \ }$ valid for $\left| t \right| \lt 1$
Therefore, the function $\dfrac{1}{{1 + 2x}}$ has the Maclaurin series $\mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{2^n}{x^n}$.
