If the coefficients of $$x2$$ and $$x3$$ are both zero, in the expansion of the $$expression1+ax+bx2(1$$−$$3x)15$$ in powers of x then the ordered pair (a$$, $$b) is equal to

Question

If the coefficients of $$x2$$ and $$x3$$ are both zero, in the expansion of the $$expression1+ax+bx2(1$$−$$3x)15$$ in powers of x then the ordered pair (a$$, $$b) is equal to

Infinity Learn · Accepted Answer

Given expression is $$1+ax+bx2(1$$−$$3x)15$$ in the expansion of binomial (1$$−$$3x)15$$ the $$(r+1)th$$ term is $$Tr+1=15Cr($$−$$3x)r=15Cr($$−$$3)rxrNow$$, coefficient of $$x2$$, in the expansion $$of1+ax+bx2(1$$−$$3x)15$$ $$15C2($$−$$3)2+a15C1($$−$$3)1+b15C0($$−$$3)0=0$$  $$(given)⇒ (105$$×$$9)−$$45a+b=0$$⇒$$45a$$−$$b=945$$  …..(1)similarly$$, the" coefficient of $$x3$$, in the expansion $$of1+ax+bx2(1$$−$$3x)15$$ is  $$15C3($$−$$3)3+a15C2($$−$$3)2+b15C1($$−$$3)1=0$$⇒    −$$12285+945a$$−$$45b$$    $$=0$$⇒        $$63a$$−$$3b=819$$⇒$$21a$$−b    $$=273----iFrom$$ Eqs. $$(i) and (ii), we get $$24a=672$$⇒$$a=28so$$, $$b=315$$⇒(a$$,$$b)=(28$$,$$315)

If the coefficients of x² and x³ are both zero, in the expansion of the expression $(1 + ax + {bx}^{2}) (1 - 3 x)^{15}$ in powers of x then the ordered pair
(a, b) is equal to

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If the coefficients of x2 and x3 are both zero, in the expansion of the expression1+ax+bx2(1−3x)15 in powers of x then the ordered pair (a, b) is equal to

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If the coefficients of x² and x³ are both zero, in the expansion of the expression $(1 + ax + {bx}^{2}) (1 - 3 x)^{15}$ in powers of x then the ordered pair
(a, b) is equal to