We call the [i]tribi [/i] of a positive integer $k$ (denoted $T(k)$) the number of all pairs $11$  in the binary representation of $k$. e.g $$T(1)=T(2)=0,\, T(3)=1, \,T(4)=T(5)=0,\,T(6)=1,\,T(7)=2.$$
Calculate $S_n=\sum_{k=1}^{2^n}T(K)$.
1. **Define the problem and initial conditions:**
   We need to calculate \( S_n = \sum_{k=1}^{2^n} T(k) \), where \( T(k) \) is the number of pairs "11" in the binary representation of \( k \). We are given the initial values:
   \[
   T(1) = T(2) = 0, \quad T(3) = 1, \quad T(4) = T(5) = 0, \quad T(6) = 1, \quad T(7) = 2
   \]

2. **Rewriting the sum:**
   We can rewrite the sum as:
   \[
   S_n = \sum_{k=1}^{2^n} T(k) = \sum_{k=0}^{2^n-1} T(k)
   \]
   since \( T(0) = T(2^n) = 0 \).

3. **Understanding the formation of \( S_{n+1} \):**
   \( S_{n+1} \) is formed by taking \( T(0), T(1), T(2), \ldots, T(2^n-1) \) and adding \( 2^n \) to each of the inputs. This is equivalent to adding a '1' to the beginning of the \( n \)-digit number in binary.

4. **Analyzing the effect of adding \( 2^n \):**
   For half of these \( k \), where the first digit of the \( n \)-digit \( k \) in binary is '1', \( T(k + 2^n) = T(k) + 1 \). Otherwise, \( T(k + 2^n) = T(k) \).

5. **Formulating the recurrence relation:**
   Therefore, we have:
   \[
   S_{n+1} = 2S_n + 2^{n-1}
   \]
   This is because for half of the \( k \), \( T(k + 2^n) = T(k) + 1 \), contributing an additional \( 2^{n-1} \) to the sum.

6. **Deriving the recurrence relation for \( S_{n+2} \):**
   Using the recurrence relation:
   \[
   S_{n+2} = 2S_{n+1} + 2^n
   \]
   and substituting \( S_{n+1} = 2S_n + 2^{n-1} \):
   \[
   S_{n+2} = 2(2S_n + 2^{n-1}) + 2^n = 4S_n + 2^n + 2^n = 4S_n + 2^{n+1}
   \]

7. **Simplifying the recurrence relation:**
   We can rewrite the recurrence relation as:
   \[
   S_{n+2} = 4S_{n+1} - 4S_n
   \]

8. **Base cases:**
   We are given:
   \[
   S_1 = 0, \quad S_2 = 1
   \]

9. **Inductive hypothesis:**
   Assume for \( n \geq 2 \):
   \[
   S_n = 2^{n-2}(n-1)
   \]
   and
   \[
   S_{n+1} = 2^{n-1}n
   \]

10. **Inductive step:**
    We need to show:
    \[
    S_{n+2} = 2^n(n+1)
    \]
    Using the recurrence relation:
    \[
    S_{n+2} = 4S_{n+1} - 4S_n
    \]
    Substituting the inductive hypothesis:
    \[
    S_{n+2} = 4(2^{n-1}n) - 4(2^{n-2}(n-1)) = 2^n \cdot 2n - 2^n \cdot (n-1) = 2^n (2n - (n-1)) = 2^n (n+1)
    \]
    Thus, the inductive step holds.

11. **Conclusion:**
    By induction, we have proved that for \( n \geq 2 \):
    \[
    S_n = 2^{n-2}(n-1)
    \]

The final answer is \( \boxed{ S_n = 2^{n-2}(n-1) } \) for \( n \geq 2 \).