If x=∑n=1∞(0.12)nx = \sum_{n=1}^{\infty} (\frac{0.1}{2})^nx=∑n=1∞(20.1)n, find the value of xxx as a simplified fraction.
119\frac{1}{19}191
120\frac{1}{20}201
19\frac{1}{9}91
118\frac{1}{18}181