The Fibonacci Sequence is defined as follows: $F_0 = 0$, $F_1 = 1$, and $F_n = F_{n-1} + F_{n-2}$ for integers $n \ge 2$. The sum $$S =\sum^{\infty}_{n=0}\frac{F^2_n}{9^n}$$
can be written as $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
1. **Expressing \( F_n \) in terms of the closed form:**
   The Fibonacci sequence \( F_n \) can be expressed using Binet's formula:
   \[
   F_n = \frac{\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n}{\sqrt{5}}
   \]
   Squaring both sides, we get:
   \[
   F_n^2 = \left( \frac{\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n}{\sqrt{5}} \right)^2
   \]
   Simplifying, we obtain:
   \[
   F_n^2 = \frac{\left(\frac{1+\sqrt{5}}{2}\right)^{2n} + \left(\frac{1-\sqrt{5}}{2}\right)^{2n} - 2 \left(\frac{1+\sqrt{5}}{2}\right)^n \left(\frac{1-\sqrt{5}}{2}\right)^n}{5}
   \]
   Since \(\left(\frac{1+\sqrt{5}}{2}\right) \left(\frac{1-\sqrt{5}}{2}\right) = -1/4\), the cross term simplifies to:
   \[
   F_n^2 = \frac{\left(\frac{1+\sqrt{5}}{2}\right)^{2n} + \left(\frac{1-\sqrt{5}}{2}\right)^{2n} + 2 \cdot \left(-\frac{1}{4}\right)^n}{5}
   \]
   \[
   F_n^2 = \frac{\left(\frac{1+\sqrt{5}}{2}\right)^{2n} + \left(\frac{1-\sqrt{5}}{2}\right)^{2n} + 2 \cdot (-1)^n \cdot \left(\frac{1}{4}\right)^n}{5}
   \]

2. **Setting up the sum \( S \):**
   We need to find:
   \[
   S = \sum_{n=0}^{\infty} \frac{F_n^2}{9^n}
   \]
   Substituting the expression for \( F_n^2 \):
   \[
   S = \sum_{n=0}^{\infty} \frac{\frac{\left(\frac{1+\sqrt{5}}{2}\right)^{2n} + \left(\frac{1-\sqrt{5}}{2}\right)^{2n} + 2 \cdot (-1)^n \cdot \left(\frac{1}{4}\right)^n}{5}}{9^n}
   \]
   Simplifying the fraction inside the sum:
   \[
   S = \frac{1}{5} \sum_{n=0}^{\infty} \left( \frac{\left(\frac{1+\sqrt{5}}{2}\right)^{2n}}{9^n} + \frac{\left(\frac{1-\sqrt{5}}{2}\right)^{2n}}{9^n} + \frac{2 \cdot (-1)^n \cdot \left(\frac{1}{4}\right)^n}{9^n} \right)
   \]
   \[
   S = \frac{1}{5} \sum_{n=0}^{\infty} \left( \left(\frac{1+\sqrt{5}}{6}\right)^{2n} + \left(\frac{1-\sqrt{5}}{6}\right)^{2n} + 2 \cdot (-1)^n \cdot \left(\frac{1}{36}\right)^n \right)
   \]

3. **Separating the sums:**
   \[
   S = \frac{1}{5} \left( \sum_{n=0}^{\infty} \left(\frac{1+\sqrt{5}}{6}\right)^{2n} + \sum_{n=0}^{\infty} \left(\frac{1-\sqrt{5}}{6}\right)^{2n} + 2 \sum_{n=0}^{\infty} (-1)^n \left(\frac{1}{36}\right)^n \right)
   \]

4. **Evaluating each geometric series:**
   For the first series:
   \[
   \sum_{n=0}^{\infty} \left(\frac{1+\sqrt{5}}{6}\right)^{2n} = \frac{1}{1 - \left(\frac{1+\sqrt{5}}{6}\right)^2}
   \]
   \[
   \left(\frac{1+\sqrt{5}}{6}\right)^2 = \frac{1 + 2\sqrt{5} + 5}{36} = \frac{6 + 2\sqrt{5}}{36} = \frac{1 + \sqrt{5}}{18}
   \]
   \[
   \sum_{n=0}^{\infty} \left(\frac{1+\sqrt{5}}{6}\right)^{2n} = \frac{1}{1 - \frac{1 + \sqrt{5}}{18}} = \frac{18}{17 - \sqrt{5}}
   \]

   For the second series:
   \[
   \sum_{n=0}^{\infty} \left(\frac{1-\sqrt{5}}{6}\right)^{2n} = \frac{1}{1 - \left(\frac{1-\sqrt{5}}{6}\right)^2}
   \]
   \[
   \left(\frac{1-\sqrt{5}}{6}\right)^2 = \frac{1 - 2\sqrt{5} + 5}{36} = \frac{6 - 2\sqrt{5}}{36} = \frac{1 - \sqrt{5}}{18}
   \]
   \[
   \sum_{n=0}^{\infty} \left(\frac{1-\sqrt{5}}{6}\right)^{2n} = \frac{1}{1 - \frac{1 - \sqrt{5}}{18}} = \frac{18}{17 + \sqrt{5}}
   \]

   For the third series:
   \[
   2 \sum_{n=0}^{\infty} (-1)^n \left(\frac{1}{36}\right)^n = 2 \cdot \frac{1}{1 + \frac{1}{36}} = 2 \cdot \frac{36}{37} = \frac{72}{37}
   \]

5. **Combining the results:**
   \[
   S = \frac{1}{5} \left( \frac{18}{17 - \sqrt{5}} + \frac{18}{17 + \sqrt{5}} + \frac{72}{37} \right)
   \]

6. **Simplifying the fractions:**
   Using the identity for the sum of fractions with conjugates:
   \[
   \frac{18}{17 - \sqrt{5}} + \frac{18}{17 + \sqrt{5}} = 18 \left( \frac{1}{17 - \sqrt{5}} + \frac{1}{17 + \sqrt{5}} \right) = 18 \left( \frac{(17 + \sqrt{5}) + (17 - \sqrt{5})}{(17 - \sqrt{5})(17 + \sqrt{5})} \right)
   \]
   \[
   = 18 \left( \frac{34}{17^2 - (\sqrt{5})^2} \right) = 18 \left( \frac{34}{289 - 5} \right) = 18 \left( \frac{34}{284} \right) = 18 \left( \frac{17}{142} \right) = \frac{306}{142}
   \]

   Combining all terms:
   \[
   S = \frac{1}{5} \left( \frac{306}{142} + \frac{72}{37} \right)
   \]

7. **Finding a common denominator and adding:**
   \[
   \frac{306}{142} = \frac{153}{71}, \quad \frac{72}{37} = \frac{144}{74}
   \]
   \[
   S = \frac{1}{5} \left( \frac{153}{71} + \frac{144}{74} \right)
   \]

8. **Simplifying the final sum:**
   \[
   S = \frac{1}{5} \left( \frac{153 \cdot 74 + 144 \cdot 71}{71 \cdot 74} \right) = \frac{1}{5} \left( \frac{11322 + 10224}{5254} \right) = \frac{1}{5} \left( \frac{21546}{5254} \right) = \frac{1}{5} \left( \frac{10773}{2627} \right) = \frac{10773}{13135}
   \]

   Simplifying further:
   \[
   S = \frac{10773}{13135} = \frac{207}{220}
   \]

The final answer is \( \boxed{207 + 220 = 427} \)