Let $n$ be a fixed positive integer. For integer $m$ satisfying $|m| \le n$, define $S_m=\displaystyle\sum_{\substack{i-j=m \\0 \le i,j \le n}} \dfrac{1}{2^{i+j}}$. Then \[\lim_{n\to \infty} \left( S_{-n}^2 + S_{-n+1}^2 +... + S_n^2 \right)\] can be expressed in the form $\dfrac{p}{q}$ for relatively prime positive integers $p,q$. Compute $100p+q$.

[i]Proposed by Vincent Huang[/i]
1. **Symmetry and Initial Computation**:
   By symmetry, \( S_{-m} = S_m \) for all \( m \). So, assuming that \( m \ge 0 \), we compute:
   \[
   S_m = \sum_{\substack{i-j=m \\ 0 \le i,j \le n}} \frac{1}{2^{i+j}}
   \]
   For \( i - j = m \), we have \( i = j + m \). Thus,
   \[
   S_m = \sum_{j=0}^{n-m} \frac{1}{2^{(j+m) + j}} = \sum_{j=0}^{n-m} \frac{1}{2^{2j+m}} = \frac{1}{2^m} \sum_{j=0}^{n-m} \frac{1}{4^j}
   \]
   The sum \(\sum_{j=0}^{n-m} \frac{1}{4^j}\) is a geometric series with the first term \(a = 1\) and common ratio \(r = \frac{1}{4}\):
   \[
   \sum_{j=0}^{n-m} \frac{1}{4^j} = \frac{1 - \left(\frac{1}{4}\right)^{n-m+1}}{1 - \frac{1}{4}} = \frac{1 - \frac{1}{4^{n-m+1}}}{\frac{3}{4}} = \frac{4}{3} \left(1 - \frac{1}{4^{n-m+1}}\right)
   \]
   Therefore,
   \[
   S_m = \frac{1}{2^m} \cdot \frac{4}{3} \left(1 - \frac{1}{4^{n-m+1}}\right) = \frac{4}{3 \cdot 2^m} \left(1 - \frac{1}{4^{n-m+1}}\right)
   \]

2. **Square of \( S_m \)**:
   \[
   S_m^2 = \left(\frac{4}{3 \cdot 2^m} \left(1 - \frac{1}{4^{n-m+1}}\right)\right)^2 = \frac{16}{9 \cdot 2^{2m}} \left(1 - \frac{1}{4^{n-m+1}}\right)^2
   \]
   Expanding the square,
   \[
   S_m^2 = \frac{16}{9 \cdot 2^{2m}} \left(1 - 2 \cdot \frac{1}{4^{n-m+1}} + \frac{1}{4^{2(n-m+1)}}\right) = \frac{16}{9 \cdot 2^{2m}} \left(1 - \frac{2}{4^{n-m+1}} + \frac{1}{4^{2(n-m+1)}}\right)
   \]

3. **Summation of \( S_m^2 \)**:
   \[
   \sum_{m=0}^n S_m^2 = \sum_{m=0}^n \frac{16}{9 \cdot 2^{2m}} \left(1 - \frac{2}{4^{n-m+1}} + \frac{1}{4^{2(n-m+1)}}\right)
   \]
   We split the sum into three parts:
   \[
   \sum_{m=0}^n S_m^2 = \frac{16}{9} \sum_{m=0}^n \frac{1}{2^{2m}} - \frac{32}{9} \sum_{m=0}^n \frac{1}{2^{2m} \cdot 4^{n-m+1}} + \frac{16}{9} \sum_{m=0}^n \frac{1}{2^{2m} \cdot 4^{2(n-m+1)}}
   \]

4. **Evaluating Each Sum**:
   - The first sum:
     \[
     \sum_{m=0}^n \frac{1}{2^{2m}} = \sum_{m=0}^n \frac{1}{4^m} = \frac{1 - \left(\frac{1}{4}\right)^{n+1}}{1 - \frac{1}{4}} = \frac{4}{3} \left(1 - \frac{1}{4^{n+1}}\right)
     \]
   - The second sum:
     \[
     \sum_{m=0}^n \frac{1}{2^{2m} \cdot 4^{n-m+1}} = \frac{1}{4^{n+1}} \sum_{m=0}^n \frac{1}{4^m} = \frac{1}{4^{n+1}} \cdot \frac{4}{3} \left(1 - \frac{1}{4^{n+1}}\right) = \frac{4}{3 \cdot 4^{n+1}} \left(1 - \frac{1}{4^{n+1}}\right)
     \]
   - The third sum:
     \[
     \sum_{m=0}^n \frac{1}{2^{2m} \cdot 4^{2(n-m+1)}} = \frac{1}{4^{2(n+1)}} \sum_{m=0}^n \frac{1}{4^m} = \frac{1}{4^{2(n+1)}} \cdot \frac{4}{3} \left(1 - \frac{1}{4^{n+1}}\right) = \frac{4}{3 \cdot 4^{2(n+1)}} \left(1 - \frac{1}{4^{n+1}}\right)
     \]

5. **Combining the Results**:
   \[
   \sum_{m=0}^n S_m^2 = \frac{16}{9} \cdot \frac{4}{3} \left(1 - \frac{1}{4^{n+1}}\right) - \frac{32}{9} \cdot \frac{4}{3 \cdot 4^{n+1}} \left(1 - \frac{1}{4^{n+1}}\right) + \frac{16}{9} \cdot \frac{4}{3 \cdot 4^{2(n+1)}} \left(1 - \frac{1}{4^{n+1}}\right)
   \]
   Simplifying each term:
   \[
   \sum_{m=0}^n S_m^2 = \frac{64}{27} \left(1 - \frac{1}{4^{n+1}}\right) - \frac{128}{27 \cdot 4^{n+1}} \left(1 - \frac{1}{4^{n+1}}\right) + \frac{64}{27 \cdot 4^{2(n+1)}} \left(1 - \frac{1}{4^{n+1}}\right)
   \]

6. **Taking the Limit as \( n \to \infty \)**:
   As \( n \to \infty \), the terms involving \( \frac{1}{4^{n+1}} \) and \( \frac{1}{4^{2(n+1)}} \) approach zero:
   \[
   \lim_{n \to \infty} \sum_{m=0}^n S_m^2 = \frac{64}{27}
   \]

7. **Final Summation**:
   Since \( S_{-m} = S_m \), we have:
   \[
   \lim_{n \to \infty} \left( S_{-n}^2 + S_{-n+1}^2 + \ldots + S_n^2 \right) = 2 \sum_{m=1}^n S_m^2 + S_0^2
   \]
   \[
   = 2 \cdot \frac{64}{27} + \frac{16}{9} = \frac{128}{27} + \frac{48}{27} = \frac{176}{27}
   \]

The final answer is \( \boxed{ 100p + q = 100 \cdot 176 + 27 = 17627 } \).