Question

If $X = \{8^n - 7n - 1 | n \in N\}$ and $Y = \{49n - 49 | n \in N\}$. Then

• A. $X \subset Y$
• B. $Y \subset X$
• C. $X = Y$
• D. $X \cap Y = \phi$

Answer 1

Answer: (A)

Given, $X = \left\{8^{n} - 7n -1 : n \in N\right\}$ and
$Y = \left\{49n - 49 : n \in N\right\}$
Now, $8^{n} - 7n - 1 = \left(7+ 1\right)^{n} - 7n - 1$
$= 7^{n}+\,{}^{n}C_{1}7^{n-1} + \,{}^{n}C_{2}7^{n-2} + ... + \,{}^{n}C_{n-2} 7^{2}$
$+\,{}^{n}C_{n-1} 7+\,{}^{n}C_{n} - 7n-1$
[$\because$ by binomial expansion, $\left(x + 1\right)^{n} = x^{n} +\,{}^{n}C_{1}x^{n-1}$
$+\,{}^{n}C_{2}x^{n-2}+... +\,{}^{n}C_{n-2}x^{2}+\,{}^{n}C_{n-1}x+\,{}^{n}C_{n}$]
$= \,{}^{n}C_{n}7^{n}+\,{}^{n}C_{1}7^{n-1}+\,{}^{n}C_{2}7^{n-2}+ ...+\,{}^{n}C_{2}7^{2}+7n+1-7n-1$
$\left(\because \,{}^{n}C_{n} =\,{}^{n}C_{0} = \,1, \,{}^{n}C_{n-1} = \,{}^{n}C_{1}=n,\,{}^{n}C_{n-2} = \,{}^{n}C_{2}\right)$
$= 7^{n}\,{}^{n}C_{n}+\,{}^{n}C_{1}7^{n-1}+\,{}^{n}C_{2}7^{n-2}+ ... +\,{}^{n}C_{2}7^{2}$
$=\,{}^{n}C_{2}7^{2}+... +\,{}^{n}C_{2}7^{n-2}+\,{}^{n}C_{1}7^{n-1}+\,{}^{n}C_{n}7^{n}$
$= 7^{2}\left[^{n}C_{2}+...+\,{}^{n}C_{n}7^{n-2}\right]$
$= 49\left[^{n}C_{2}+...+\,{}^{n}C_{n}7^{n-2}\right]$
$\therefore 8^{n} - 7n - 1$ is a multiple of $49$ for $n \ge 2$
For $n = 1, 8^{n}-7n - 1 = 8 - 8 = 0$
$\therefore 8^{n}- 7n- 1$ is a multiple of $49$ for all $n \in N$.
$X$ contains elements which are multiples of $49$ and clearly
$Y$ contains all multiples of $49$.
$\therefore X \subset Y$