: Sum of the series S=3−1 10C0−32   10C1+32 10C2−33 10C3+….…+310 10C10 is 

: Sum of the series S=31 10C032   10C1+32 10C233 10C3+....+310 10C10 is 

  1. A

    29

  2. B

    210-1

  3. C

    132112

  4. D

    13(210)

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    Solution:

    We have 

    S=31 10C010C1(3)+10C23210C333+.....+10C10310 =13(13)10=13210

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