Question:
If $X=\left\{8^{n}-7 n-1: n \in N\right\}$ and $Y=\{49 n-49: n \in N\}$. Then,
(a) $X \subset Y$
(b) $Y \subset X$
(c) $X=Y$
(d) $X \cap Y=\phi$
Solution:
$X=\left\{8^{n}-7 n-1, n \in N\right\}$
$Y=\{49 n-49 ; n \in N\}$
Since $8^{n}-7 n-1=(1+7)^{n}-7 n-1$
$=\left(1+7 n+{ }^{n} C_{2} 7^{2}+{ }^{n} C_{3} 7^{3}+\ldots+7^{n}\right)-7 n-1$
$=1+7 n+{ }^{n} C_{2} 49+{ }^{n} C_{3} 7^{3}+\ldots+7^{n}-7 n-1$
$=7^{2}\left({ }^{n} C_{2}+{ }^{n} C_{3} 7+\ldots+7^{n-2}\right)$
$=49\left({ }^{n} C_{2}+{ }^{n} C_{3} 7+\ldots+7^{n-2}\right)$
i.e. $8^{n}-7 n-1$ is a multiple of 49
i. e. $X \subset Y$
Here $X=\{0,49,490,4067, \ldots\}$
$Y=\{0,49,98, \ldots \ldots\}$
$Y \not \subset X($ since $98 \notin x)$
$\therefore X \subset Y$
Hence, the correct answer is option A.