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. $X\cap Y=\varphi$

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 $49$.
$\therefore X\subset Y$