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Question

nr=0nCr(2rn)2=
  1. n.2n+1
  2. n.2n
  3. (n1)2n
  4. n22n


Solution

The correct option is B n.2n
nr=0nCr(2rn)2=nr=0(4r2+n24rn)nCr=nr=04r2nCr+n2nr=0nCr4nnr=0nCr=n.2n

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