Answer
$1-\frac{3}{4}x-3\Sigma_{n=2}^{\infty}\frac{5.9.13......(4n-7)x^{n}}{4^{n}n!}$
and
$R=1$
Work Step by Step
${(1-x)^{3/4}}=(1+(-x))^{3/4}$
$=1-\frac{3}{4}x-3\Sigma_{n=2}^{\infty}\frac{5.9.13......(4n-7)x^{n}}{4^{n}n!}$
$$\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{5.9.13......(4n-7)(4(n+1)-7)x^{n+1}}{4^{n+1}(n+1)!}}{\frac{5.9.13......(4n-7)x^{n}}{4^{n}n!}}|$$
$=|x|$
The series will converge when $|x|\lt 1$ so $R=1$.
