For a positive integer $n$, let $s(n)$ denote the sum of the binary digits of $n$. Find the sum $s(1)+s(2)+s(3)+...+s(2^k)$ for each positive integer $k$.
1. **Define the sum \( S_k \):**
   \[
   S_k = s(1) + s(2) + \cdots + s(2^k)
   \]

2. **Establish the recurrence relation:**
   - Note that \( s(2^a) = 1 \) for any integer \( a \).
   - For any \( x < 2^a \), \( s(2^a + x) = s(x) + 1 \).

3. **Express \( S_{k+1} \) in terms of \( S_k \):**
   \[
   S_{k+1} = \sum_{i=1}^{2^{k+1}} s(i)
   \]
   Split the sum into two parts:
   \[
   S_{k+1} = \sum_{i=1}^{2^k} s(i) + \sum_{i=1}^{2^k} s(2^k + i)
   \]
   Using the property \( s(2^k + i) = s(i) + 1 \):
   \[
   S_{k+1} = \sum_{i=1}^{2^k} s(i) + \sum_{i=1}^{2^k} (s(i) + 1)
   \]
   Simplify the second sum:
   \[
   S_{k+1} = \sum_{i=1}^{2^k} s(i) + \sum_{i=1}^{2^k} s(i) + \sum_{i=1}^{2^k} 1
   \]
   \[
   S_{k+1} = 2 \sum_{i=1}^{2^k} s(i) + 2^k
   \]
   Since \( \sum_{i=1}^{2^k} s(i) = S_k \):
   \[
   S_{k+1} = 2S_k + 2^k
   \]

4. **Base case:**
   \[
   S_1 = s(1) + s(2) = 1 + 1 = 2
   \]

5. **Inductive hypothesis:**
   Assume \( S_k = 2^{k-1}k + 1 \) holds for some \( k \).

6. **Inductive step:**
   Show \( S_{k+1} = 2^k(k+1) + 1 \):
   \[
   S_{k+1} = 2S_k + 2^k
   \]
   Substitute the inductive hypothesis \( S_k = 2^{k-1}k + 1 \):
   \[
   S_{k+1} = 2(2^{k-1}k + 1) + 2^k
   \]
   Simplify:
   \[
   S_{k+1} = 2^k k + 2 + 2^k
   \]
   Combine like terms:
   \[
   S_{k+1} = 2^k k + 2^k + 2
   \]
   Factor out \( 2^k \):
   \[
   S_{k+1} = 2^k (k + 1) + 1
   \]
   This completes the inductive step.

7. **Conclusion:**
   By induction, we have shown that:
   \[
   S_k = 2^{k-1}k + 1
   \]

The final answer is \( \boxed{ 2^{k-1}k + 1 } \)