If `A = {x : x = 6^n - 5n - 1, n in N}` and `B = {x : x = 25(n - 1), n in N}`, then

Options

A. A = B
B. `B sub A`
C. `A sube B`
D. `B sube A`

Answer

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

1. **Set A**: \( A = \{ x : x = 6^n - 5n - 1, n \in \mathbb{N} \} \)
2. **Set B**: \( B = \{ x : x = 25(n - 1), n \in \mathbb{N} \} \)

### Step 1: Calculate elements of Set A
We will compute the first few elements of set \( A \) by substituting natural numbers for \( n \).

- For \( n = 1 \):
\[ x = 6^1 - 5 \cdot 1 - 1 = 6 - 5 - 1 = 0 \]

- For \( n = 2 \):
\[ x = 6^2 - 5 \cdot 2 - 1 = 36 - 10 - 1 = 25 \]

- For \( n = 3 \):
\[ x = 6^3 - 5 \cdot 3 - 1 = 216 - 15 - 1 = 200 \]

Thus, the first few elements of set \( A \) are:
\[ A = \{ 0, 25, 200, \ldots \} \]

### Step 2: Calculate elements of Set B
Now, we will compute the first few elements of set \( B \) by substituting natural numbers for \( n \).

- For \( n = 1 \):
\[ x = 25(1 - 1) = 25 \cdot 0 = 0 \]

- For \( n = 2 \):
\[ x = 25(2 - 1) = 25 \cdot 1 = 25 \]

- For \( n = 3 \):
\[ x = 25(3 - 1) = 25 \cdot 2 = 50 \]

Thus, the first few elements of set \( B \) are:
\[ B = \{ 0, 25, 50, 75, 100, \ldots \} \]

### Step 3: Compare Sets A and B
Now we will compare the elements of sets \( A \) and \( B \):

- Elements of \( A \): \( 0, 25, 200, \ldots \)
- Elements of \( B \): \( 0, 25, 50, 75, 100, \ldots \)

From our calculations, we can see that:
- \( 0 \) is in both \( A \) and \( B \).
- \( 25 \) is in both \( A \) and \( B \).
- However, \( 50 \) is in \( B \) but not in \( A \).
- \( 200 \) is in \( A \) but not in \( B \).

### Conclusion
Since every element of \( B \) is not contained in \( A \) (specifically, \( 50 \) is in \( B \) but not in \( A \)), we conclude that:
- \( A \) is not equal to \( B \).
- \( B \) is not a subset of \( A \).
- However, \( A \) is a subset of \( B \) because all elements of \( A \) are present in \( B \).

Thus, the correct answer is:
\[ A \subset B \]