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Question: If the coefficient of \[{{x}^{2}}\And {{x}^{3}}\] are both zero in the expansion of the expression,\...

If the coefficient of x2&x3{{x}^{2}}\And {{x}^{3}} are both zero in the expansion of the expression,(1+ax+bx2)(1−3x)15(1+ax+b{{x}^{2}}){{(1-3x)}^{15}} in powers of x , then the ordered pair of (a,b) is equal to:
(a). (28,315)
(b). (-54,315)
(c). (-21,714)
(d). (24,861)

Explanation

Solution

The main idea of the problem is try to find all combinations of the coefficients of x2&x3{{x}^{2}}\And {{x}^{3}} and equate them to zero to get two equation of the unknown “a” and “b”. Next we will solve those equations to get the desired ordered pair. In order to expand the 2nd2^{nd} expression we need the standard formula of Binomial Theorem.

Complete step by step answer:
For any two number” a & b” (real or complex) we have,
(a+b)n=an+nC1an−1b+nC2an−2b2+...+bn........(1){{(a+b)}^{n}}={{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{{b}^{n}}........(1)
We will expand the 2nd2^{nd} expression according to this formula,
(1−3x)15=1−15C1(3x)+15C2(9x2)−15C3(27x3)+....+(−3x)15{{(1-3x)}^{15}}=1-{}^{15}{{C}_{1}}(3x)+{}^{15}{{C}_{2}}(9{{x}^{2}})-{}^{15}{{C}_{3}}(27{{x}^{3}})+....+{{(-3x)}^{15}}
After simplifying the above expression, we get,
(1−3x)15=1−45x+945x2−12285x3+...−14348907x15{{(1-3x)}^{15}}=1-45x+945{{x}^{2}}-12285{{x}^{3}}+...-14348907{{x}^{15}} …….(2)
Now we will compute the all possible combinations of x2&x3{{x}^{2}}\And {{x}^{3}} keeping in mind the product rule of algebraic expressions.
First we try to get the equation by assigning the coefficients of x2{{x}^{2}} to be zero.
We will avoid unnecessary term to avoid ambiguity.
(1+ax+bx2)(1−45x+945x2−12285x3) (1+ax+b{{x}^{2}})(1-45x+945{{x}^{2}}-12285{{x}^{3}})\text{ }
The coefficient of x2{{x}^{2}}will be ,
945-45a+b=0\text{945-45a+b=0}……………………..(3)
The coefficient of x3{{x}^{3}} will be zero,
945a−45b=12285945a-45b=12285………………………..(4)
We have to solve the equations (3) & (4) to get the value of “a” & “b”.
First we will divide both side of (4) by 45:
21a−b=27321a-b=273……………….(5)
We will add the equation (3) and (5) to get value of “a”:
45a−21a=67245a-21a=672
⇒a=67224=28\Rightarrow a=\dfrac{672}{24}=28
So, a=28 , to get the value of b , we simply substitute the value of a to (5) to get b:
b =(21)28-273=315
This is our desired ordered pair (28,315).

So, the correct answer is “Option A”.

Note: Please try to be patient to calculate the product of equation (2). As wrong calculations might lead to wrong answers. Never change the order of the 2nd2^{nd} expression, If you do , you might get the wrong answer.