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Question

Let E=(2n+1)(2n+3)(2n+5)(4n3)(4n1) where n>1, then 2n E is divisible by


A

2nCn

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B

4nC2n

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C

2nCn2

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D

4nCn2

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Solution

The correct option is B

4nC2n


E=(2n+1)(2n+3)(2n+5)(4n3)(4n1)E=(2n)!(2n+1)(2n+2)(2n+3)(2n+4)(2n+5)(4n1)4n(2n)!(2n+2)(2n+4)(4n)E=(4n)!n!(2n)!n!2n(n+1)(n+2)(2n)E=(4n)!n!(2n)!(2n)!2n2nE=n!4nC2n
Hence 2nE is divisible by 4nC2n


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