If A = {5^{n} - 4n - 1 : n in N} and B = {16( | vedclass

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(C) We have $A = 5^{n} - 4n - 1 = (1 + 4)^{n} - 4n - 1$. Using the binomial expansion, $(1 + 4)^{n} = {}^{n}C_{0} + {}^{n}C_{1}(4) + {}^{n}C_{2}(4^{2}

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$A \subseteq B$

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(C) We have $A = 5^{n} - 4n - 1 = (1 + 4)^{n} - 4n - 1$. Using the binomial expansion, $(1 + 4)^{n} = {}^{n}C_{0} + {}^{n}C_{1}(4) + {}^{n}C_{2}(4^{2}) + \dots + {}^{n}C_{n}(4^{n})$. So, $A = (1 + 4n + 16({}^{n}C_{2} + {}^{n}C_{3}(4) + \dots + {}^{n}C_{n}(4^{n-2}))) - 4n - 1$. $A = 16({}^{n}C_{2} + {}^{n}C_{3}(4) + \dots + {}^{n}C_{n}(4^{n-2}))$. This shows that every element of $A$ is a multiple of $16$. For $n=1$, $5^{1}-4(1)-1 = 0$. For $n=2$, $5^{2}-4(2)-1 = 16$. For $n=3$, $5^{3}-4(3)-1 = 112 = 16 	imes 7$. Thus, $A = \{0, 16, 112, \dots\}$. $B = \{16(n-1) : n \in N\} = \{0, 16, 32, 48, \dots\}$. Since every element of $A$ is a multiple of $16$, $A \subseteq B$.

If $A = \{5^{n} - 4n - 1 : n \in N\}$ and $B = \{16(n - 1) : n \in N\}$, then:

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