The people in an infinitely long line are numbered $1,2,3,\dots$. Then, each person says either ``Karl" or ``Lark" independently and at random. Let $S$ be the set of all positive integers $i$ such that people $i$, $i+1$, and $i+2$ all say ``Karl," and define $X = \sum_{i \in S} 2^{-i}$. Then the expected value of $X^2$ is $\frac{m}{n}$, where $m$ and $n$ are positive integers with $\gcd(m,n)=1$. Compute $100m+n$.

[i]Proposed by Ankit Bisain[/i]
1. **Define the problem and variables:**
   - Let \( S \) be the set of all positive integers \( i \) such that people \( i \), \( i+1 \), and \( i+2 \) all say "Karl."
   - Define \( X = \sum_{i \in S} 2^{-i} \).
   - We need to find the expected value of \( X^2 \).

2. **Calculate the probability \( \mathbb{P}(i \in S) \):**
   - Each person says "Karl" or "Lark" independently with probability \( \frac{1}{2} \).
   - The probability that three consecutive people say "Karl" is \( \left(\frac{1}{2}\right)^3 = \frac{1}{8} \).
   - Therefore, \( \mathbb{P}(i \in S) = \frac{1}{8} \).

3. **Calculate the probability \( \mathbb{P}(i \text{ and } j) \):**
   - For \( j = i+1 \), the events \( i \in S \) and \( j \in S \) are not independent.
     - The probability that people \( i, i+1, i+2, i+3 \) all say "Karl" is \( \left(\frac{1}{2}\right)^4 = \frac{1}{16} \).
     - Therefore, \( \mathbb{P}(i \text{ and } i+1) = \frac{1}{16} \).
   - For \( j = i+2 \), the events \( i \in S \) and \( j \in S \) are not independent.
     - The probability that people \( i, i+1, i+2, i+3, i+4 \) all say "Karl" is \( \left(\frac{1}{2}\right)^5 = \frac{1}{32} \).
     - Therefore, \( \mathbb{P}(i \text{ and } i+2) = \frac{1}{32} \).
   - For \( j \geq i+3 \), the events \( i \in S \) and \( j \in S \) are independent.
     - The probability that people \( i, i+1, i+2 \) say "Karl" and people \( j, j+1, j+2 \) say "Karl" is \( \left(\frac{1}{2}\right)^6 = \frac{1}{64} \).
     - Therefore, \( \mathbb{P}(i \text{ and } j) = \frac{1}{64} \).

4. **Calculate \( \mathbb{E}[X^2] \):**
   \[
   \mathbb{E}[X^2] = \mathbb{E}\left[\left(\sum_{i \in S} 2^{-i}\right)^2\right]
   \]
   \[
   = \mathbb{E}\left[\sum_{i \in S} 2^{-2i}\right] + 2 \mathbb{E}\left[\sum_{i, j \in S \land i < j} 2^{-i-j}\right]
   \]
   \[
   = \sum_{i=1}^\infty 2^{-2i} \mathbb{P}(i) + 2 \sum_{i=1}^\infty \sum_{j=i+1}^\infty 2^{-i-j} \mathbb{P}(i \text{ and } j)
   \]

5. **Evaluate the sums:**
   \[
   \sum_{i=1}^\infty 2^{-2i} \mathbb{P}(i) = \sum_{i=1}^\infty 2^{-2i} \cdot \frac{1}{8} = \frac{1}{8} \sum_{i=1}^\infty 2^{-2i} = \frac{1}{8} \cdot \frac{1}{3} = \frac{1}{24}
   \]
   \[
   2 \sum_{i=1}^\infty \left(2^{-2i-1} \mathbb{P}(i \text{ and } i+1) + 2^{-2i-2} \mathbb{P}(i \text{ and } i+2) + \sum_{j=i+3}^\infty 2^{-i-j} \mathbb{P}(i \text{ and } j)\right)
   \]
   \[
   = 2 \sum_{i=1}^\infty \left(2^{-2i-1} \cdot \frac{1}{16} + 2^{-2i-2} \cdot \frac{1}{32} + \sum_{j=i+3}^\infty 2^{-i-j} \cdot \frac{1}{64}\right)
   \]
   \[
   = 2 \sum_{i=1}^\infty \left(2^{-2i} \cdot \frac{1}{32} + 2^{-2i} \cdot \frac{1}{128} + 2^{-2i} \cdot \frac{1}{256}\right)
   \]
   \[
   = 2 \sum_{i=1}^\infty 2^{-2i} \left(\frac{1}{32} + \frac{1}{128} + \frac{1}{256}\right)
   \]
   \[
   = 2 \cdot \frac{1}{3} \left(\frac{1}{32} + \frac{1}{128} + \frac{1}{256}\right)
   \]
   \[
   = \frac{2}{3} \left(\frac{1}{32} + \frac{1}{128} + \frac{1}{256}\right)
   \]
   \[
   = \frac{2}{3} \left(\frac{8}{256} + \frac{2}{256} + \frac{1}{256}\right)
   \]
   \[
   = \frac{2}{3} \cdot \frac{11}{256} = \frac{22}{768} = \frac{11}{384}
   \]

6. **Combine the results:**
   \[
   \mathbb{E}[X^2] = \frac{1}{24} + \frac{11}{384} = \frac{16}{384} + \frac{11}{384} = \frac{27}{384} = \frac{9}{128}
   \]

The final answer is \( \boxed{ 100m + n = 100 \times 9 + 128 = 1028 } \).