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

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(B) We have $4^n - 3n - 1 = (1 + 3)^n - 3n - 1$. Using the binomial expansion,$(1 + 3)^n = 1 + n(3) + \frac{n(n-1)}{2}(3^2) + \dots + 3^n$. So,$4^n -

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$Y$

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(B) We have $4^n - 3n - 1 = (1 + 3)^n - 3n - 1$. Using the binomial expansion,$(1 + 3)^n = 1 + n(3) + \frac{n(n-1)}{2}(3^2) + \dots + 3^n$. So,$4^n - 3n - 1 = 1 + 3n + 9 \binom{n}{2} + 27 \binom{n}{3} + \dots + 3^n - 3n - 1 = 9 \left[ \binom{n}{2} + 3 \binom{n}{3} + \dots + 3^{n-2} \right]$. This shows that $4^n - 3n - 1$ is a multiple of $9$ for all $n \ge 2$. For $n = 1$,$4^1 - 3(1) - 1 = 0$. For $n = 2$,$4^2 - 3(2) - 1 = 9$. Thus,$X = \{ 0, 9, 27, 54, \dots \}$. For $Y = \{ 9(n-1) : n \in N \}$,we have $Y = \{ 0, 9, 18, 27, 36, 45, 54, \dots \}$. Since every element of $X$ is a multiple 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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