Solution

First we consider the fact that $7R_{n} +2R_{n-1}-9R_{n-2}= 64$

Now, consider the fact that

$S= \frac{R_0}{1} +\frac{R_1}{2} +\frac{R_2}{4} + \frac{R_3}{8} ...$

Thus,

$7S= \frac{7R_0}{1} +\frac{7R_1}{2} +\frac{7R_2}{4} + \frac{7R_3}{8}...$

and

$S= \frac{2R_0}{2} +\frac{2R_1}{4} +\frac{2R_2}{8} + \frac{2R_3}{16} ...$

and

$\frac{-9S}{4}= \frac{-9R_0}{4} +\frac{-9R_1}{8} +\frac{-9R_2}{16} + \frac{-9R_3}{32} ...$

Adding these together we get that

$S=\frac{420}{23}$

Since 420 and 23 are relatively prime we find that the answer is $\boxed{443}$

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