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Question

If the number of terms in the expansion of (12x+4x2)n, x0, is 28, then the sum of the coefficients of all the terms in this expansions, is

A
64
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B
2187
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C
243
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D
729
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Solution

The correct option is D 729

Clearly, number of terms in the expansion of
(12x+4x2)n is (n+2)(n+1)2 or n+2C2
[assuming 1x and 1x2 distinct]
(n+2)(n+1)2=28
(n + 2)(n + 1) = 56 =( 6 + 1) (6 + 2) n = 6
Hence, sum of coefficients =(12+4)6=36=729


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