Step-by-step Solution:

Let \( x_n = 4^n - 3n - 1 \), where \( n \in \mathbb{N} \). 1. For \( n = 1 \): \[ x_1 = 4 - 3 - 1 = 0. \] 2. For \( n \geq 2 \): Using the expansion \( 4^n = (1 + 3)^n \), we get: \[ x_n = (1 + 3)^n - 3n - 1. \] Applying the **Binomial Theorem**, \[ \begin{aligned} x_n &= {^nC_0} + {^nC_1} \cdot 3 + {^nC_2} \cdot 3^2 + {^nC_3} \cdot 3^3 + \ldots + {^nC_n} \cdot 3^n - 3n - 1 \\ &= 1 + 3n + {^nC_2} \cdot 3^2 + {^nC_3} \cdot 3^3 + \ldots + {^nC_n} \cdot 3^n - 3n - 1. \end{aligned} \] Simplifying further, \[ x_n = 9 \left[ {^nC_2} + 3 \cdot {^nC_3} + 3^2 \cdot {^nC_4} + \ldots + 3^{n-2} \cdot {^nC_n} \right]. \] Thus, for \( n \geq 2 \), \( x_n \) is a positive integral multiple of 9. \[\] The Sets \( X \) and \( Y \) \[\] 1. Set \( X \): \( X \) consists of all positive integral multiples of 9 of the form: \[ x_n = 9 \left[{^nC_2} + 3 \cdot {^nC_3} + 3^2 \cdot {^nC_4} + \ldots + 3^{n-2} \cdot {^nC_n}\right], \] along with \( x_1 = 0 \). Explicitly, \[ X = \{0, 9, 54, 243, \ldots\}. \] 2. Set \( Y \): Let \( Y = \{9(n-1): n \in \mathbb{N}\} \). This means \( Y \) consists of all integral multiples of 9, including 0. Explicitly, \[ Y = \{0, 9, 18, 27, 36, 54, \ldots\}. \] Conclusion Clearly, \[ X \subset Y. \]