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Question

The coefficient of x7in the expansion of (1xx2+x3)6 is

A
-132
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B
-144
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C
132
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D
144
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Solution

The correct option is B -144
(1xx2+x3)6=[(1x)x2(1x)]6=(1x)6(1x2)6=(16x+15x220x3+15x46x5+x6)×(16x2+15x420x6+15x86x10+x12)Coefficient of x7=(6)(20)+(20)(15)+(6)(6)=144

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