A) X
B) Y
C) N
D) None of these
Correct Answer: B
Solution :
Since, \[{{4}^{n}}-3n-1={{(3+1)}^{n}}-3n-1\] \[={{3}^{n}}{{+}^{n}}{{C}_{1}}{{3}^{n-1}}{{+}^{n}}{{C}_{2}}{{3}^{n-2}}+.....{{+}^{n}}{{C}_{n-1}}3{{+}^{n}}{{C}_{n}}-3n-1\] \[{{=}^{n}}{{C}_{2}}{{3}^{2}}{{+}^{n}}{{C}_{3}}{{.3}^{3}}+...{{+}^{n}}{{C}_{n}}{{3}^{n}}\],\[n\,(H\cap B)=64,\,\,n(B\cap C)=80\] \[=9{{[}^{n}}{{C}_{2}}{{+}^{n}}{{C}_{3}}(3)+.....{{+}^{n}}{{C}_{n}}{{3}^{n-1}}]\] \[\therefore \]\[{{4}^{n}}-3n-1\] is a multiple of 9 for \[n\ge 2\]. For \[n=1,\] \[{{4}^{n}}-3n-1\] = \[4-3-1=0\], For \[n=2,\] \[{{4}^{n}}-3n-1\]= \[16-6-1=9\] \[\therefore \]\[{{4}^{n}}-3n-1\] is a multiple of 9 for all \[n\in N\] \[\therefore \] X contains elements, which are multiples of 9, and clearly Y contains all multiples of 9. \[\therefore \]\[X\subseteq Y\] i.e., \[X\cup Y=Y\].
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