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Question

If X=8n7n1,nN and Y=49(n1),nN, then -

A
X ⊆ Y
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B
Y ⊆ X
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C
X = Y
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D
None of these
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Solution

The correct option is A X ⊆ Y
Conventional Approach:

Since 8n7n1=(7+1)n7n1
=7n+nC17n1+nC27n2++nCn17+nCn7n1
=nC272+nC373++nCn7n,(nC0=nCn,nC1=nCn1etc.)
=49[nC2+nC3(7)++nCn7n2]
8n7n1 is a multiple of 49 for n2
For n=1,8n7n1=871=0;
For n=2,8n7n1=64141=49
8n7n1 is a multiple of 49 for all nN.
X contains elements which are multiples of 49 and Y clearly contains all multiples of 49.
XY.

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.
XY.

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