• # question_answer If $X=\{{{8}^{n}}-7n-1:n\in N\}$ and $Y=\{49(n-1):n\in N\},$ then A) $X\subseteq Y$ B) $Y\subseteq X$ C) $X=Y$ D) None of these

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. $\therefore$ $X\subseteq Y$.