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

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

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$Y$

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(B) For set $X$,we have $X = \{ 4^n - 3n - 1 : n \in N \}$.For $n = 1$,$4^1 - 3(1) - 1 = 0$.For $n = 2$,$4^2 - 3(2) - 1 = 16 - 6 - 1 = 9$.For $n = 3$,$4^3 - 3(3) - 1 = 64 - 9 - 1 = 54$.Thus,$X = \{ 0, 9, 54, 243, \dots \}$.For set $Y$,we have $Y = \{ 9(n - 1) : n \in N \}$.For $n = 1$,$9(1 - 1) = 0$.For $n = 2$,$9(2 - 1) = 9$.For $n = 3$,$9(3 - 1) = 18$.Thus,$Y = \{ 0, 9, 18, 27, \dots \}$.Since every element of $X$ is a multiple of $9$ and $X \subset Y$ (as $4^n - 3n - 1$ is always divisible by $9$ for $n \in N$),the union $X \cup Y$ is equal to $Y$.

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

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