IfA=\{ (4(n-1); n∈ N\} and B= (4-4n:ne Mwhere N is the set of natural numbers, then whichone is true?8.(1) A c B(2) BcA(3) AnB=A

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X = {4^{n} - 3 n - 1 : n \in N}

Y = {9 (n - 1) : n \in N}

N

X \cup Y

X = {4^{n} - 3 n - 1 : n \in N}

Y = {9 (n - 1) : n \in N}

X \cup Y

For natural number n, $(n!)^{2}$ > $n^{n}$ , if

(n!)^{2} > n^{n}

X = {4^{n} - 3 n - 1 : n ϵ N} a n d Y = {9 (n - 1) : n ϵ N}

X \cup Y

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