Question

If X = $\{8^n-7n-1:n\in N\}andY=\left\{49\right(n-1):n\in N\}$ , then

• A.
• B.
• C. X = Y
• D. None of these

Answer 1

The correct option is A

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

= $7^n+{}^n{C}_1{7}^{n-1}+{}^n{C}_2{7}^{n-2}+.....+{}^n{C}_{n-1}7+{}^n{C}_n-7n-1$

= ${}^n{C}_2{7}^2+{}^n{C}_3{7}^3+...+{}^n{C}_n{7}^n,({}^n{C}_0={}^n{C}_n{}^n{C}_1={}^n{C}_{n-1}etc,)$

= 49[${}^n{C}_2+{}^n{C}_3\left(7\right)+........+{}^n{C}_n{7}^{n-2}]$

$\therefore$ $8^n-7n-1$ 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 n $\ge$ N

$\therefore$ X contains elements which are multiples of 49 and clearly $\gamma$

contains all multiplies of 49. $\therefore$ X $\subseteq$ Y.