If `X={8^n-7n-1 | n in N}` and `Y={49n-49| n in N}`. Then

Options

A. `X sub Y`
B. `Y sub X`
C. `X = Y`
D. `X nn Y = phi`

Explanation

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

1. **Set \( X \)**: \( X = \{ 8^n - 7n - 1 \mid n \in \mathbb{N} \} \)
2. **Set \( Y \)**: \( Y = \{ 49n - 49 \mid n \in \mathbb{N} \} \)

### Step 1: Calculate the elements of set \( X \)

We will compute the elements of set \( X \) for the first few natural numbers \( n \):

- For \( n = 1 \):
\[ 8^1 - 7 \cdot 1 - 1 = 8 - 7 - 1 = 0 \]

- For \( n = 2 \):
\[ 8^2 - 7 \cdot 2 - 1 = 64 - 14 - 1 = 49 \]

- For \( n = 3 \):
\[ 8^3 - 7 \cdot 3 - 1 = 512 - 21 - 1 = 490 \]

Thus, the first few elements of set \( X \) are:
\[ X = \{ 0, 49, 490, \ldots \} \]

### Step 2: Calculate the elements of set \( Y \)

Next, we compute the elements of set \( Y \) for the first few natural numbers \( n \):

- For \( n = 1 \):
\[ 49 \cdot 1 - 49 = 49 - 49 = 0 \]

- For \( n = 2 \):
\[ 49 \cdot 2 - 49 = 98 - 49 = 49 \]

- For \( n = 3 \):
\[ 49 \cdot 3 - 49 = 147 - 49 = 98 \]

Thus, the first few elements of set \( Y \) are:
\[ Y = \{ 0, 49, 98, \ldots \} \]

### Step 3: Compare the sets \( X \) and \( Y \)

Now we will compare the elements of both sets:

- The elements of \( X \) are \( 0, 49, 490, \ldots \)
- The elements of \( Y \) are \( 0, 49, 98, \ldots \)

From our calculations, we can observe that:
- Every element of \( X \) (i.e., \( 0, 49, 490 \)) is present in \( Y \).
- However, \( Y \) contains elements like \( 98 \) which are not present in \( X \).

### Conclusion

Since every element of \( X \) is in \( Y \) but not vice versa, we conclude that:
\[ X \subseteq Y \]