If `A={4^(n)-3n-1:n in N) and B={9(n-1):n in N}`
then

Options

A. `B sub A`
B. `A cup B=N`
C. `A subB`
D. None of these

Explanation

To determine whether the set \( B \) is a subset of the set \( A \), we need to analyze the definitions of both sets.

Given:
- \( A = \{ 4^n - 3n - 1 : n \in \mathbb{N} \} \)
- \( B = \{ 9(n - 1) : n \in \mathbb{N} \} \)

### Step 1: Analyze the elements of set \( B \)
The elements of set \( B \) can be expressed as:
\[ B = \{ 9(n - 1) : n \in \mathbb{N} \} \]
This means for \( n = 1, 2, 3, \ldots \):
- For \( n = 1 \): \( 9(1 - 1) = 0 \)
- For \( n = 2 \): \( 9(2 - 1) = 9 \)
- For \( n = 3 \): \( 9(3 - 1) = 18 \)
- For \( n = 4 \): \( 9(4 - 1) = 27 \)
- And so on...

Thus, \( B = \{ 0, 9, 18, 27, \ldots \} \).

### Step 2: Analyze the elements of set \( A \)
The elements of set \( A \) can be expressed as:
\[ A = \{ 4^n - 3n - 1 : n \in \mathbb{N} \} \]
We will calculate a few values of \( A \):
- For \( n = 1 \): \( 4^1 - 3(1) - 1 = 4 - 3 - 1 = 0 \)
- For \( n = 2 \): \( 4^2 - 3(2) - 1 = 16 - 6 - 1 = 9 \)
- For \( n = 3 \): \( 4^3 - 3(3) - 1 = 64 - 9 - 1 = 54 \)
- For \( n = 4 \): \( 4^4 - 3(4) - 1 = 256 - 12 - 1 = 243 \)

Thus, the first few elements of \( A \) are \( \{ 0, 9, 54, 243, \ldots \} \).

### Step 3: Check if \( B \subseteq A \)
Now we need to check if every element of \( B \) is also in \( A \):
- The element \( 0 \) is in both \( A \) and \( B \).
- The element \( 9 \) is also in both \( A \) and \( B \).
- The next element of \( B \) is \( 18 \), which is not in \( A \) (as we found \( 54 \) for \( n=3 \)).
- Continuing, \( 27 \) is also not in \( A \).

Since \( 18 \) and \( 27 \) from \( B \) are not found in \( A \), we conclude that \( B \) is not a subset of \( A \).

### Conclusion
Thus, we can conclude that:
\[ B \nsubseteq A \]