To solve the problem, we need to analyze the two sets \( A \) and \( B \) given in the question.
**Step 1: Define the sets.**
- Set \( A \) is defined as \( A = \{ x : x = 3^n - 2n - 1, n \in \mathbb{N} \} \)
- Set \( B \) is defined as \( B = \{ x : x = 4(n-1), n \in \mathbb{N} \} \)
**Step 2: Calculate elements of set \( A \) for \( n = 1, 2, 3, \ldots \)**
- For \( n = 1 \):
\[
x = 3^1 - 2 \cdot 1 - 1 = 3 - 2 - 1 = 0
\]
- For \( n = 2 \):
\[
x = 3^2 - 2 \cdot 2 - 1 = 9 - 4 - 1 = 4
\]
- For \( n = 3 \):
\[
x = 3^3 - 2 \cdot 3 - 1 = 27 - 6 - 1 = 20
\]
- For \( n = 4 \):
\[
x = 3^4 - 2 \cdot 4 - 1 = 81 - 8 - 1 = 72
\]
So, the first few elements of set \( A \) are \( \{ 0, 4, 20, 72, \ldots \} \).
**Step 3: Calculate elements of set \( B \) for \( n = 1, 2, 3, \ldots \)**
- For \( n = 1 \):
\[
x = 4(1-1) = 4 \cdot 0 = 0
\]
- For \( n = 2 \):
\[
x = 4(2-1) = 4 \cdot 1 = 4
\]
- For \( n = 3 \):
\[
x = 4(3-1) = 4 \cdot 2 = 8
\]
- For \( n = 4 \):
\[
x = 4(4-1) = 4 \cdot 3 = 12
\]
So, the first few elements of set \( B \) are \( \{ 0, 4, 8, 12, \ldots \} \).
**Step 4: Compare the elements of sets \( A \) and \( B \)**
- From the calculated elements, we can see:
- \( 0 \in A \) and \( 0 \in B \)
- \( 4 \in A \) and \( 4 \in B \)
- \( 20 \in A \) but \( 20 \notin B \)
- \( 72 \in A \) but \( 72 \notin B \)
The elements of set \( A \) include numbers like \( 20 \) and \( 72 \) which are not in set \( B \).
**Step 5: Determine the relationship between sets \( A \) and \( B \)**
- Since not all elements of \( A \) are in \( B \), we conclude that \( A \) is not a subset of \( B \).
- However, since \( 0 \) and \( 4 \) are common in both sets, and \( B \) contains multiples of \( 4 \), we can see that \( A \) does contain some elements that are also in \( B \).
**Final Conclusion:**
- The correct relationship is that \( A \) is not a subset of \( B \) and \( B \) is not a subset of \( A \). Therefore, the correct option is that \( A \) and \( B \) share some elements but neither is a subset of the other.
