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Question

nN,4n3n1 is divisible by

A
3
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B
8
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C
9
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D
11
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Solution

The correct options are
A 3
C 9
4n3n1=(1+3)n3n1=(1+nC13+nC232+.........nCn3n)3n1
=(1+3n+nC232+........)3n1=nC232+nC333+............+nCn3n
=32(nC2+nC33+...........+nCn3n2)=9× Integer.
Hence given number is divisible by 9. Hence it is also divisible by 3.

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