If X

Solution:

Given:

$X=\left\{8^{n}-7 n-1: n \in N\right\}$ and $Y=\{49(n-1): n \in N\}$

To prove:

$X \subseteq Y$

Let:

$x_{n}=8^{n}-7 n-1, n \in N$

$\Rightarrow x_{1}=8-7-1=0$

For any $n \geqslant 2$, we have :

$x_{n}=8^{n}-7 n-1=(1+7)^{n}-7 n-1$

$\Rightarrow x_{n}={ }^{n} C_{0}+{ }^{n} C_{1} \cdot 7+{ }^{n} C_{2} \cdot 7^{2}+{ }^{n} C_{3} \cdot 7^{3}+\ldots+{ }^{n} C_{n} \cdot 7^{n}-7 n-1$

$\Rightarrow x_{n}=1+7 n+{ }^{n} C_{2} \cdot 7^{2}+{ }^{n} C_{3} \cdot 7^{3}+\ldots+7^{n}-7 n-1 \quad\left[\because{ }^{n} C_{0}=1\right.$ and $\left.{ }^{n} C_{1}=n\right]$

$\Rightarrow x_{n}=7^{2}\left\{{ }^{n} C_{2}+{ }^{n} C_{3} .7+{ }^{n} C_{4} 7^{2}+\ldots+{ }^{n} C_{n} \cdot 7^{n-2}\right\}$

$\Rightarrow x_{n}=49\left\{{ }^{n} C_{2}+{ }^{n} C_{3} \cdot 7+{ }^{n} C_{4} 7^{2}+\ldots+{ }^{n} C_{n} \cdot 7^{n-2}\right\}$

Thus, $x_{n}$ is some positive integral multiple of 49 for all $n \geqslant 2$.

$X$ consists of all those positive integral multiple $s$ of 49 that are of the form $49\left\{{ }^{n} C_{2}+{ }^{n} C_{3} \cdot 7+{ }^{n} C_{4} 7^{2}+\ldots+{ }^{n} C_{n} \cdot 7^{n-2}\right\}$ along with zero.

$Y=\{49(n-1): n \in N\}$ implies that it consists of all integral multiple $s$ of 49 along with zero.

$\therefore X \subseteq Y$