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If $\; (X =4^{n} -3n-1) : n \in N \;$ and $\;Y=9(n-1) : n \in N\;$ where N is the set of natural numbers then $\;X \cup Y\;$ is equal to :

$(a)\;X\qquad(b)\;Y\qquad(c)\;N\qquad(d)\;Y-X$

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