SATHEE: Binomial Theorem 1 Question 2

Binomial Theorem 1 Question 2

2.

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

(2019 Main, 10 April I)

(a) $(28, 315)$

(b) $(- 21, 714)$

(d) $(- 54, 315)$

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Answer:

Correct Answer: 2. (a)

Solution:

Given expression is $(1 + a x + b x^{2}) (1 - 3 x)^{15}$ . In the expansion of binomial $(1 - 3 x)^{15}$ , the $(r + 1)$ th term is

$T_{r + 1} =^{15} C_{r} (- 3 x)^{r} =^{15} C_{r} (- 3)^{r} x^{r}$

Now, coefficient of $x^{2}$ , in the expansion of $(1 + a x + b x^{2}) (1 - 3 x)^{15}$ is

$^{15} C_{2} (- 3)^{2} + a^{15} C_{1} (- 3)^{1} + b^{15} C_{0} (- 3)^{0} = 0$ (given)

$\Rightarrow (105 \times 9) - 45 a + b = 0$

$\Rightarrow 45 a - b = 945$ …….(i)

Similarly, the coefficient of $x^{3}$ , in the expansion of $(1 + a x + b x^{2}) (1 - 3 x)^{15}$ is

$^{15} C_{3} (- 3)^{3} + a^{15} C_{2} (- 3)^{2} + b^{15} C_{1} (- 3)^{1} = 0$ (given)

$\Rightarrow - 12285 + 945 a - 45 b = 0$

$\Rightarrow 63 a - 3 b = 819$

$\Rightarrow 21 a - b = 273$ …….(ii)

From Eqs. (i) and (ii), we get

$24 a = 672 \Rightarrow a = 28$

So, $b = 315$

$\Rightarrow (a, b) = (28, 315)$

Binomial Theorem 1 Question 2

2.

Answer:

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Binomial Theorem 1 Question 2

2.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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