Answer
Maclaurin series is: $\Sigma_{n=0}^{\infty}(n+1)x^{n}$
and $R=1$
Work Step by Step
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{{(n+2)}x^{n+1}}{(n+1)x^{n}}|$
$=\lim\limits_{n \to\infty}|(\frac{n+2}{n+1}).x|$
$=|x|\lt 1$
Maclaurin series is: $\Sigma_{n=0}^{\infty}(n+1)x^{n}$
and $R=1$
