If (1+x)n=C0+C1x+C2x2+…+Cnxn  then C02+C12+C22+C32+…+Cn2 is equal to

 If (1+x)n=C0+C1x+C2x2++Cnxn  then C02+C12+C22+C32++Cn2 is equal to

  1. A

    n!n!n!

  2. B

    (2n)!n!n!

  3. C

    (2n)!n!

  4. D

    None of these 

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    Solution:

    we have ,(1+x)n=C0+C1x+C2x2++Cnxn-----(i)

     and 1+1xn=C0+C11x+C21x2++Cn1xn(ii)

    on multiplying Eqs. (i) and (ii) and taking the
    coefficient of constant terms in right hand side

    =C02+C12+C22++Cn2

    In right hand side (1+x)n1+1xn or in 1xn(1+x)2n or 

    term containing xn in (1 + x)2n.

    clearly, the coefficient of xn in (1 + x)2n is equal to

     2nCn=(2n)!n!n!

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