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Question

For all $$n\in N,{ 2 }^{ 4n }-15n-1$$ is divisibl;e by


A
225
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B
125
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C
325
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D
None of these
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Solution

The correct option is D $$225$$
We have $${ 2 }^{ 4n }={ \left( { 2 }^{ 4 } \right)  }^{ n }={ \left( 16 \right)  }^{ n }={ \left( 1+15 \right)  }^{ n }$$
$$\therefore { 2 }^{ 4n }=1+^{ n }{ { C }_{ 1 } }\times 15+^{ n }{ { C }_{ 2 } }\times { 15 }^{ 2 }+^{ n }{ { C }_{ 3 } }\times { 15 }^{ 3 }+...\\ \Rightarrow { 2 }^{ 4n }-1-15n={ 15 }^{ 2 }\left[ ^{ n }{ { C }_{ 2 } }+^{ n }{ { C }_{ 3 } }\times 15+... \right] $$
$$=225K,$$ where $$K$$ is an integer.
Hence $${ 2 }^{ 4n }-1-15n$$ is divisible by $$225$$.

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