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Question

If the coefficients of x3 and x4 in the expansion (1+ax+bx2)(12x)18 in powers of x are both zero, then (a,b) is equal to

A
(16,2723)
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B
(16,2513)
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C
(14,2513)
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D
(14,2723)
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Solution

The correct option is A (16,2723)
This can be written as
(12x)18+ax(12x)18+bx2(12x)18
Hence coefficient of x4
=18C424a18C323+b18C222
=0
Hence 18C422a18C321+b18C2=0 ...(i)
The coefficient of x3
=18C323+a18C222b18C121=0
18C322+a18C221b18C1=0 ..(ii)
Solving the above equations, we get
a=16 and b=2723

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