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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
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B
XY
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C
YX
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D
None of these
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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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