Answer
Maclaurin's series is: $\Sigma_{n=0}^{\infty}\frac{{(ln2})^{n}x^{n}}{(n)!}$
and $R=\infty$
Work Step by Step
$f(x)=2^{x}=\Sigma_{n=0}^{\infty}\frac{(ln2)^{n}x^{n}}{(n)!}$
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{(ln2)^{n+1}x^{n+1}}{(n+1)!}}{\frac{(ln2)^{n}x^{n}}{(n)!}}|$
$=\lim\limits_{n \to\infty}|\frac{(ln2).x}{n+1}|$
$=|x|\lt 1$
Therefore, the Maclaurin's series converges for all values of $x$.
Maclaurin's series is: $\Sigma_{n=0}^{\infty}\frac{{(ln2})^{n}x^{n}}{(n)!}$
and $R=\infty$
