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Question

If 2nC2+2nC4+2nC6++ 2nC2n=511, then absolue value of nC0nC131+nC232++(1)n ncn3n is

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Solution

We know that 2nC0+ 2nC2+ 2nC4+ 2nC6++2nC2n=22n1512=22n1n=5

Let S=nC0nC131+nC232++(1)n nCn3n
=nr=0(1)r nCr3r
=nr=0nCr(3)r =(13)n=(2)n
Since, n=5
S=(2)5=32
|S|=32

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