Question

If $X=8^n-7n-1,n\in NandY=49(n-1),n\in N$, then -

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

Answer 1

The correct option is A X ⊆ Y

Conventional Approach:

Since $8^n-7n-1=(7+1{)}^n-7n-1$
$=7^n{+}^n{C}_1{7}^{n-1}{+}^n{C}_2{7}^{n-2}+\cdots {+}^n{C}_{n-1}7{+}^n{C}_n-7n-1$
${=}^n{C}_2{7}^2{+}^n{C}_3{7}^3+\cdots {+}^n{C}_n{7}^n,{(}^n{C}_0{=}^n{C}_n{,}^n{C}_1{=}^n{C}_{n-1}etc.)$
$=49{[}^n{C}_2{+}^n{C}_3\left(7\right)+\cdots {+}^n{C}_n{7}^{n-2}]$
$\therefore 8^n-7n-1$ 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 all $n\in N.$
$\therefore$ X contains elements which are multiples of 49 and Y clearly contains all multiples of 49.
$\therefore X\subset Y.$

Best Approach:

Put n = 1, X = 0, Y = 0
For n = 2, X = 49, Y = 49
For n = 3, X = 490, Y = 98
$\therefore$ X contains elements which are multiples of 49 and Y clearly contains all multiples of 49.
$\therefore X\subset Y.$