If X = {4^n - 3n - 1 : n in N} and Y = {9(n - | vedclass

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(A) Given $X = \{4^n - 3n - 1 : n \in N\}$ and $Y = \{9(n - 1) : n \in N\}$.By binomial expansion,$4^n = (1 + 3)^n = 1 + n(3) + \frac{n(n-1)}{2!} (3^2

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(A) Given $X = \{4^n - 3n - 1 : n \in N\}$ and $Y = \{9(n - 1) : n \in N\}$.By binomial expansion,$4^n = (1 + 3)^n = 1 + n(3) + \frac{n(n-1)}{2!} (3^2) + \dots + 3^n$.So,$4^n - 3n - 1 = 1 + 3n + \frac{9n(n-1)}{2} + \dots - 3n - 1 = \frac{9n(n-1)}{2} + \dots = 9 \left[ \frac{n(n-1)}{2} + \dots \right]$.This shows that every element of $X$ is a multiple of $9$,and since $n(n-1)$ is always even,$\frac{n(n-1)}{2}$ is an integer.Thus,$X \subseteq Y$.Alternatively,testing values:For $n=1, X = \{4^1 - 3(1) - 1\} = \{0\}$.For $n=2, X = \{4^2 - 3(2) - 1\} = \{16 - 6 - 1\} = \{9\}$.For $n=3, X = \{4^3 - 3(3) - 1\} = \{64 - 9 - 1\} = \{54\}$.$X = \{0, 9, 54, \dots\}$.$Y = \{0, 9, 18, 27, 36, 45, 54, \dots\}$.Since all elements of $X$ are in $Y$,$X \cap Y = X$.

If $X = \{4^n - 3n - 1 : n \in N\}$ and $Y = \{9(n - 1) : n \in N\}$,then $X \cap Y =$

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