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Question

The sum of the coefficients of all the integral powers of x in (1+3x)100 is


A

299 [2100 - 1]

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B

299 [2100 + 1]

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C

4100 + 2100

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D

None of these

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Solution

The correct option is B

299 [2100 + 1]


100C0 + 100C232 + 100C434 .... + 100C1003100 = 12[(3+1)100+(31)100]=12[4100+2100]


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