A) \[X\subseteq Y\]
B) \[Y\subseteq X\]
C) \[X=Y\]
D) None of these
Correct Answer: A
Solution :
Since \[{{8}^{n}}-7n-1={{(7+1)}^{n}}-7n-1\] \[={{7}^{n}}{{+}^{n}}{{C}_{1}}{{7}^{n-1}}{{+}^{n}}{{C}_{2}}{{7}^{n-2}}+.....{{+}^{n}}{{C}_{n-1}}7{{+}^{n}}{{C}_{n}}-7n-1\] \[{{=}^{n}}{{C}_{2}}{{7}^{2}}{{+}^{n}}{{C}_{3}}{{7}^{3}}+..{{+}^{n}}{{C}_{n}}{{7}^{n}}\],\[{{(}^{n}}{{C}_{0}}{{=}^{n}}{{C}_{n}},{{\,}^{n}}{{C}_{1}}{{=}^{n}}{{C}_{n-1}}\,\text{etc}\text{.)}\] \[=49{{[}^{n}}{{C}_{2}}{{+}^{n}}{{C}_{3}}(7)+......{{+}^{n}}{{C}_{n}}{{7}^{n-2}}]\] \[\therefore \] \[{{8}^{n}}-7n-1\] is a multiple of 49 for \[n\ge 2\] For \[n=1\], \[{{8}^{n}}-7n-1=8-7-1=0\]; For \[n=2,\] \[{{8}^{n}}-7n-1=64-14-1=49\] \[\therefore \] \[{{8}^{n}}-7n-1\] is a multiple of 49 for all \[n\in N.\] \[\therefore \] X contains elements which are multiples of 49 and clearly Y contains all multiplies of 49. \[\therefore \] \[X\subseteq Y\].
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