Let an=∑r=0n 1 nCr2 and bn=∑r=0n (n−r) nCr2then bn equals 

Let an=r=0n1 nCr2 and bn=r=0n(nr) nCr2

then bn equals 

  1. A

    an

  2. B

    nan

  3. C

    n2an

  4. D

    0

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    Solution:

    We have

    bn=r=0nnr nCr2=r=0nnr nCnr2=r=0nr nCr2

    Thus, 

    bn+bn=r=0nnrnCr2+r=0nrnCr2=nr=0n1 nCr2=nan

     bn=n2an

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