if (1+x)15=C0+C1x+C2x2+…+C15x15 then C2+2C3+3C4+…+14C15 is equal to 

if (1+x)15=C0+C1x+C2x2++C15x15 then C2+2C3+3C4++14C15 is equal to 

  1. A

    14214

  2. B

    13214+1

  3. C

    132141

  4. D

    None of these

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    Solution:

    we have (1+x)15=C0+C1x+C2x2++C15x15

     (1+x)151x=C1+C2x++C15x14

    On differentiating both sides w.r.t. x, we get

    x15(1+x)14(1+x)15+1x2=C2+2C3x++14C15x13

    On putting x = t, we get 

    C2+2C3++14C15=15214215+1=13214+1

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