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Question

# If the coefficient of x2 and x3 are both zero, in the expansion of the expression (1+ax+bx2)(1−3x)15 in powers of x, then the ordered pair (a,b) is equal to :

A
(21,714)
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B
(54,315)
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C
(28,861)
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D
(28,315)
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Solution

## The correct option is D (28,315)Coefficient of x2 in (1+ax+bx2)(1−3x)15 ⇒ 15C2(−3)2+a (15C1)(−3)+b (15C0)(−3)0=0 ⇒15⋅142⋅9−3a⋅15+b=0⇒15⋅63−45a+b=0 ⋯(1) Coefficient of x3 in (1+ax+bx2)(1−3x)15 ⇒ 15C3⋅(−3)3+a(15C2)(−3)2+b(15C1)(−3)=0 ⇒15⋅14⋅133⋅2⋅1⋅32−a⋅3⋅15⋅142+15⋅b=0⇒21⋅13−21a+b=0 ⋯(2) On subtracting (1)−(2) , we have 15⋅63−21⋅13−24a=0⇒a=28⇒b=315

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