Binomial series

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This topic contains 3 replies, has 3 voices, and was last updated by  Soumyadeep mandal 1 week, 3 days ago.

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  • #32408

    Aniruddha Bardhan
    Participant

    See the attachment.

    My approach:

    let S=20C0+20C1+…+20C10

    20C0+20C1+…+20C20=2^20

    2S-20C10=2^20

    S=2^19+.5*20C10

    How option B is correct?

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    #32468

    A.Venkat
    Participant

    Hint :

    You can try to use the identity that $$\binom{n}{0} + binom{n}{1} + …… + binom{n}{n} = 2^n $$ and then manipulate the rest.

    • This reply was modified 1 week, 5 days ago by  A.Venkat.
    #32507

    Aniruddha Bardhan
    Participant

    Ya I use this approach. But got S=2^19+ (1/2)*20C10

    #32599

    Soumyadeep mandal
    Participant

    well even I am getting the answer as 2^19 + (1/2)*(20C10)…. I think the sum is till (20C9) only, then only we can get option 4 as the answer …

     

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