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

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(B) Given $X = \{ 4^n - 3n - 1 : n \in N \}$.For $n=1, 4^1 - 3(1) - 1 = 0$.For $n=2, 4^2 - 3(2) - 1 = 16 - 6 - 1 = 9$.For $n=3, 4^3 - 3(3) - 1 = 64 -

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$Y$

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(B) Given $X = \{ 4^n - 3n - 1 : n \in N \}$.For $n=1, 4^1 - 3(1) - 1 = 0$.For $n=2, 4^2 - 3(2) - 1 = 16 - 6 - 1 = 9$.For $n=3, 4^3 - 3(3) - 1 = 64 - 9 - 1 = 54$.So,$X = \{ 0, 9, 54, \dots \}$.Given $Y = \{ 9(n - 1) : n \in N \}$.For $n=1, 9(1-1) = 0$.For $n=2, 9(2-1) = 9$.For $n=3, 9(3-1) = 18$.So,$Y = \{ 0, 9, 18, 27, \dots \}$.Since every element of $X$ is a multiple of $9$ and $X \subset Y$,it follows that $X \cup Y = Y$.

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

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