For every positive integer $n$, determine the biggest positive integer $k$ so that $2^k |\ 3^n+1$
To determine the largest positive integer \( k \) such that \( 2^k \) divides \( 3^n + 1 \) for every positive integer \( n \), we need to analyze the divisibility properties of \( 3^n + 1 \) with respect to powers of 2.

1. **Check divisibility by \( 2 \):**
   \[
   3^n + 1 \equiv 0 \pmod{2}
   \]
   Since \( 3 \equiv 1 \pmod{2} \), we have:
   \[
   3^n \equiv 1^n \equiv 1 \pmod{2}
   \]
   Therefore:
   \[
   3^n + 1 \equiv 1 + 1 \equiv 0 \pmod{2}
   \]
   This shows that \( 2^1 = 2 \) divides \( 3^n + 1 \) for any \( n \).

2. **Check divisibility by \( 4 \):**
   \[
   3^n + 1 \equiv 0 \pmod{4}
   \]
   Since \( 3 \equiv -1 \pmod{4} \), we have:
   \[
   3^n \equiv (-1)^n \pmod{4}
   \]
   Therefore:
   \[
   3^n + 1 \equiv (-1)^n + 1 \pmod{4}
   \]
   - If \( n \) is even, \( (-1)^n = 1 \):
     \[
     3^n + 1 \equiv 1 + 1 \equiv 2 \pmod{4}
     \]
     This shows that \( 3^n + 1 \) is not divisible by \( 4 \) when \( n \) is even.
   - If \( n \) is odd, \( (-1)^n = -1 \):
     \[
     3^n + 1 \equiv -1 + 1 \equiv 0 \pmod{4}
     \]
     This shows that \( 4 = 2^2 \) divides \( 3^n + 1 \) when \( n \) is odd.

3. **Check divisibility by \( 8 \):**
   \[
   3^n + 1 \equiv 0 \pmod{8}
   \]
   Since \( 3 \equiv 3 \pmod{8} \), we have:
   \[
   3^n \equiv 3^n \pmod{8}
   \]
   We need to check the values of \( 3^n \pmod{8} \) for different \( n \):
   - If \( n = 1 \):
     \[
     3^1 + 1 = 4 \equiv 4 \pmod{8}
     \]
   - If \( n = 2 \):
     \[
     3^2 + 1 = 9 + 1 = 10 \equiv 2 \pmod{8}
     \]
   - If \( n = 3 \):
     \[
     3^3 + 1 = 27 + 1 = 28 \equiv 4 \pmod{8}
     \]
   - If \( n = 4 \):
     \[
     3^4 + 1 = 81 + 1 = 82 \equiv 2 \pmod{8}
     \]
   From these calculations, we see that \( 3^n + 1 \) is never divisible by \( 8 \) for any \( n \).

Therefore, the largest \( k \) such that \( 2^k \) divides \( 3^n + 1 \) for every positive integer \( n \) is \( k = 2 \) when \( n \) is odd, and \( k = 1 \) when \( n \) is even.

The final answer is \( \boxed{ k = 2 } \) for odd \( n \) and \( k = 1 \) for even \( n \).