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 $n=3$,$4^3 - 3(3) - 1 = 64 - 9 - 1 = 54$.Using the binomial expansion,$4^n = (1+3)^n = 1 + n(3) + \frac{n(n-1)}{2}(3^2) + \dots + 3^n$.Thus,$4^n - 3n - 1 = 9 	imes [\frac{n(n-1)}{2} + \dots + 3^{n-2}]$.This shows that every element in $X$ is a multiple of $9$ (including $0$).$Y = \{ 9(n-1) : n \in N \} = \{ 0, 9, 18, 27, \dots \}$,which is the set of all non-negative multiples of $9$.Since all elements of $X$ are multiples of $9$,$X \subseteq Y$.Therefore,$X \cup Y = Y$.

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

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