Let `A = {64n , n in N}`
and `B ={3^(2n+2) - 8^(n) -9, n in N}` then

Options

A. `A sube B, A ne B`
B. `B sube A, B ne A`
C. A = B
D. `A cap B = phi`

Answer

To solve the problem, we need to analyze the sets \( A \) and \( B \) defined as follows:

- \( A = \{ 64n \mid n \in \mathbb{N} \} \)
- \( B = \{ 3^{2n+2} - 8^n - 9 \mid n \in \mathbb{N} \} \)

### Step 1: Determine the elements of set \( A \)

Set \( A \) consists of elements formed by multiplying \( 64 \) with natural numbers \( n \).

- For \( n = 1 \): \( 64 \times 1 = 64 \)
- For \( n = 2 \): \( 64 \times 2 = 128 \)
- For \( n = 3 \): \( 64 \times 3 = 192 \)

Thus, the first few elements of set \( A \) are:
\[ A = \{ 64, 128, 192, 256, \ldots \} \]

### Step 2: Determine the elements of set \( B \)

Set \( B \) is defined by the expression \( 3^{2n+2} - 8^n - 9 \).

- For \( n = 1 \):
\[ 3^{2 \times 1 + 2} - 8^1 - 9 = 3^4 - 8 - 9 = 81 - 8 - 9 = 64 \]

- For \( n = 2 \):
\[ 3^{2 \times 2 + 2} - 8^2 - 9 = 3^6 - 64 - 9 = 729 - 64 - 9 = 656 \]

- For \( n = 3 \):
\[ 3^{2 \times 3 + 2} - 8^3 - 9 = 3^8 - 512 - 9 = 6561 - 512 - 9 = 6040 \]

Thus, the first few elements of set \( B \) are:
\[ B = \{ 64, 656, 6040, \ldots \} \]

### Step 3: Compare sets \( A \) and \( B \)

From our calculations:
- Set \( A \) contains elements \( 64, 128, 192, \ldots \)
- Set \( B \) contains elements \( 64, 656, 6040, \ldots \)

### Step 4: Identify the relationship between sets \( A \) and \( B \)

1. **Common Elements**: Both sets contain the element \( 64 \).
2. **Subset Check**:
- Set \( A \) is an infinite set of multiples of \( 64 \).
- Set \( B \) grows much faster due to the exponential terms involved.

### Conclusion

Since \( B \) contains \( 64 \) and other elements that are not in \( A \), and \( A \) contains multiples of \( 64 \) (which are not necessarily in \( B \)), we can conclude that:

- \( B \) is a subset of \( A \) because all elements of \( B \) (after \( 64 \)) are not present in \( A \) and \( 64 \) is the only common element.

Thus, the correct option is:
**B is a subset of A**.