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Question

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


A

12(350+1)

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B

12(350)

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C

12(3501)

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D

12(250+1)

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Solution

The correct option is A

12(350+1)


Let Tr+1 be the general term in the expension of (12x)50
Tr+1=50Cr(1)50r(2x12)r=50Cr2rxr2(1)r
For the integral power of x, r should be even integer.
Sum of coefficients =25r=050C2r(2)2r
=12[(1+2)50+(12)50]=12(350+1)


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