The AM of  2n+1C0,2n+1C1,2n+1C2,…,2n+1Cn is

The AM of  2n+1C0,2n+1C1,2n+1C2,,2n+1Cn is

  1. A

    2nn

  2. B

    2nn+1

  3. C

    22nn

  4. D

    22n(n+1)

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    Solution:

     2n+1C0+2n+1C1+2n+1C2++2n+1C2n+2n+1C2n+1=22n+1

    Now  2n+1C0=2n+1C2n+1, 2n+1C1=2n+1C2n2n+1Cr=2n+1C2nr+1

    So, sum of first (n + 1 ) terms

    = sum of last (n + 1 ) terms

     2n+1C0+2n+1C1+2n+1C2++2n+1Cn=22n  2n+1C0+2n+1C1+2n+1C2++2n+1Cnn+1=22n(n+1)

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