If an=∑r=0n 1 nCr, then ∑r=0n r nCr  equals:

If an=r=0n1 nCr, then r=0nr nCr  equals:

  1. A

    n2an

  2. B

    n4an

  3. C

    nan

  4. D

    (n1)an

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    Solution:

    Let bn=r=0nr nCr=r=0nnr nCnr=r=0nnr nCr

     2bn=r=0nr+(nr) nCr=nr=0n1 nCr=nan

     bn=n2an

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