If X=`{4^n-3n-1:n in N}` and `{9(n-1):n in N, then precisely:

Options

A. `XsubY`
B. `XsubeY`
C. X = Y
D. `XsupeY`

Answer

To solve the problem, we need to analyze the two sets given:

1. Set \( X = \{ 4^n - 3n - 1 : n \in \mathbb{N} \} \)
2. Set \( Y = \{ 9(n-1) : n \in \mathbb{N} \} \)

### Step 1: Calculate elements of Set \( X \)

We will calculate the first few elements of Set \( X \) by substituting natural numbers for \( n \).

- For \( n = 1 \):
\[ X_1 = 4^1 - 3 \cdot 1 - 1 = 4 - 3 - 1 = 0 \]

- For \( n = 2 \):
\[ X_2 = 4^2 - 3 \cdot 2 - 1 = 16 - 6 - 1 = 9 \]

- For \( n = 3 \):
\[ X_3 = 4^3 - 3 \cdot 3 - 1 = 64 - 9 - 1 = 54 \]

- For \( n = 4 \):
\[ X_4 = 4^4 - 3 \cdot 4 - 1 = 256 - 12 - 1 = 243 \]

Thus, the first few elements of Set \( X \) are:
\[ X = \{ 0, 9, 54, 243, \ldots \} \]

### Step 2: Calculate elements of Set \( Y \)

Now, we will calculate the first few elements of Set \( Y \).

- For \( n = 1 \):
\[ Y_1 = 9(1-1) = 9 \cdot 0 = 0 \]

- For \( n = 2 \):
\[ Y_2 = 9(2-1) = 9 \cdot 1 = 9 \]

- For \( n = 3 \):
\[ Y_3 = 9(3-1) = 9 \cdot 2 = 18 \]

- For \( n = 4 \):
\[ Y_4 = 9(4-1) = 9 \cdot 3 = 27 \]

- For \( n = 5 \):
\[ Y_5 = 9(5-1) = 9 \cdot 4 = 36 \]

Thus, the first few elements of Set \( Y \) are:
\[ Y = \{ 0, 9, 18, 27, 36, \ldots \} \]

### Step 3: Compare Sets \( X \) and \( Y \)

Now we need to check if all elements of Set \( X \) are also in Set \( Y \).

- \( 0 \in Y \)
- \( 9 \in Y \)
- \( 54 \) is not explicitly listed in \( Y \), but we can see that \( 54 = 9 \cdot 6 \), which is a multiple of 9 and thus belongs to \( Y \).
- \( 243 \) is also not in the initial list, but it can be checked that it is \( 9 \cdot 27 \), which is also a multiple of 9.

Since all calculated elements of Set \( X \) are multiples of 9, we can conclude that:
\[ X \subseteq Y \]

### Conclusion

The final conclusion is that Set \( X \) is a subset of Set \( Y \).