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Question

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


Solution

We know that  2nC0+2nC2+2nC4+2nC6++2nC2n=22n1 
512=22n1
n=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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