I have the following recurrence relation (where $n$ is an integer greater than $0$):
$$\begin{array}{*{20}{l}} \begin{array}{l} {x_0} = \frac{1}{2},\\ {x_1} = \frac{3}{2},\\ {x_2} = \frac{1}{2}, \end{array}\\ {{x_{3n}} = \frac{1}{2}\cdot{x_{3n - 1}},}\\ {{x_{3n + 1}} = {x_{3n}} + 1,}\\ {{x_{3n + 2}} = \frac{1}{3}\cdot{x_{3n + 1}}} \end{array}$$
Question: how to find ${x_{146}}$ ( that is, ${{x_{3n + 2}}}$ for $n = 48$ )? If I am not mistaken, it should be equal to $$\frac{{3({6^{50}} - 1)}}{{5 \cdot {6^{50}}}}$$ or somewhere close to this, but what is the correct way to obtain this result?