The sum of coefficients of integral powers of r in the binomial expansion of (1−2x)50is 

The sum of coefficients of integral powers of r in the binomial expansion of (12x)50is 

  1. A

    12350+1

  2. B

    12350

  3. C

    123501

  4. D

    12250+1

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    Solution:

     Let Tr+1 be the general term in the expansion of (12x)50

     Tr+1=50Cr(1)50r2x1/2r=50Cr2rxn/2(1)r

    For the integral power of x, r should be even integer.

     Sum of coefficients =r=02550C2r(2)2r =12(1+2)50+(12)50=12350+1

    Aliter We have,

    (12x)50=C0C12x+C2(2x)2++C50(2x)50(i) (1+2x)50=C0+C12x+C2(2x)2++C50(2x)50 (ii

    On adding Eqs. (i) and (ii), we get

    (12x)50+(1+2x)50=2C0+C2(2x)2 ++C50(2x)50

     (12x)50+(1+2x)502=C0+C2(2x)2++C50(2)50

     (1)50+(3)502=C0+C2(2)2++C50(2)50

     1+3502=C0+C2(2)2++C50(2)50

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