Question

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

A
X=Y
B
XY
C
YX
D
None of these

Solution

## The correct option is B $$X\subset Y$$We have$${ 8 }^{ n }-7n-1={ (1+7) }^{ n }-7n-1$$$$={ _{ }^{ n }{ C } }_{ 0 }+{ _{ }^{ n }{ C } }_{ 1 }\times 7+{ _{ }^{ n }{ C } }_{ 2 }{ 7 }^{ 2 }+....+{ _{ }^{ n }{ C } }_{ n }\times { 7 }^{ n }-(7n+1)\quad$$$$={ 7 }^{ 2 }({ _{ }^{ n }{ C } }_{ 2 }+{ _{ }^{ n }{ C } }_{ 3 }\times 7+....+{ _{ }^{ n }{ C } }_{ n }\times { 7 }^{ n-2 })$$$$\therefore$$ $$X$$ contains some multiples $$49$$Clearly $$Y$$ contains all multiplies of $$49$$ including $$0$$.$$\therefore$$ $$X\subset Y$$Mathematics

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