The correct option is A X ⊆ Y
Conventional Approach:
Since 8n−7n−1=(7+1)n−7n−1
=7n+nC17n−1+nC27n−2+⋯+nCn−17+nCn−7n−1
=nC272+nC373+⋯+nCn7n,(nC0=nCn,nC1=nCn−1etc.)
=49[nC2+nC3(7)+⋯+nCn7n−2]
∴ 8n−7n−1 is a multiple of 49 for n≥2
For n=1,8n−7n−1=8−7−1=0;
For n=2,8n−7n−1=64−14−1=49
∴8n−7n−1 is a multiple of 49 for all n∈N.
∴ X contains elements which are multiples of 49 and Y clearly contains all multiples of 49.
∴X⊂Y.
Best Approach:
Put n = 1, X = 0, Y = 0
For n = 2, X = 49, Y = 49
For n = 3, X = 490, Y = 98
∴ X contains elements which are multiples of 49 and Y clearly contains all multiples of 49.
∴X⊂Y.