If X = {8ⁿ – 7n – 1 / n ∈ N} and Y = {49 (n - 1) / n ∈ N}, then

x ⊂ Y

We have,

$8^n-7 n-1=(7+1)^n-7 n-1=\left({ }^n C_2 7^2+{ }^n C_3 7^3+...+{ }^n C_n 7^n\right)$

$=49\left({ }^n C_2+{ }^n C_3 7+...+{ }^n C_n 7^{n-2}\right) \text { for } n \geq 2$

For $n=1,8^n-7^{n-1}=0$

Thus, 8ⁿ - 7n-1 is a multiple of 49 for n ≥ 2 and 0 for n = 1. Hence X consists of all positive integral multiple of 49 of the form 49 K_n, where K_n = ⁿC₂ + ⁿC₃ 7+…+ ⁿC_n 7^{n - 2} together with zero. Also Y consists of all positive integral multiple of 49 including zero. Therefore, X ⊂ Y.

Hence (1) is the correct answer.