Answer
$R=2$
and
Interval of convergence: $(-1,3]$
Work Step by Step
Given: $$a_n=\frac{(x-1)^{n}}{(2n-1)2^{n}}$$$$|\frac{a_{n+1}}{a_{n}}|=|\dfrac{\dfrac{(x-1)^{n+1}}{(2(n+1)-1)2^{n+1}}}{\dfrac{(x-1)^{n}}{(2n-1)2^{n}}}|$$
Take limits on both sides, we have $$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(x-1)^{n+1}}{(2(n+1)-1)2^{n+1}}}{\dfrac{(x-1)^{n}}{(2n-1)2^{n}}}|=\frac{x-1}{2}$$
Hence,
$R=2$
and
Interval of convergence: $(-1,3]$