Suppose [x] denote the greatest integer ≤x and n∈N, then limn→∞  nCox2+ nC1x2+⋯+ nCnx22n−2 is equal to

Suppose [x] denote the greatest integer x and nN, then 

limn nCox2+ nC1x2++ nCnx22n2 is equal to

  1. A

    12x2

  2. B

    x2

  3. C

    2x2

  4. D

    4x2

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    Solution:

    We know x1<[x]xxR, there fore  nCkx21< nCkx2nCkx2

     k=0n nCkx21<k=0n nCkx2k=0nnCkx2 x22n(n+1)2n2<12n2k=0n nCkx2x22n2n2

    Taking limit as n ,and using sandwich theorem, we get

    limn12n2k=0n nCkx2=x2

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