SATHEE: Binomial Theorem 1 Question 25

Binomial Theorem 1 Question 25

27.

If the coefficients of $x^{3}$ and $x^{4}$ in the expansion of $(1 + a x + b x^{2}) (1 - 2 x)^{18}$ in powers of $x$ are both zero, then $(a, b)$ is equal to

(a) $(16, \frac{251}{3})$

(b) $(14, \frac{251}{3})$

(d) $(16, \frac{272}{3})$

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Answer:

Correct Answer: 27. (d)

Solution:

To find the coefficient of $x^{3}$ and $x^{4}$ , use the formula of coefficient of $x^{r}$ in $(1 - x)^{n}$ is $(- 1)^{r} C_{r}$ and then simplify.

In expansion of $(1 + a x + b x^{2}) (1 - 2 x)^{18}$ .

Coefficient of $x^{3} =$ Coefficient of $x^{3}$ in $(1 - 2 x)^{18}$

$\begin{matrix} + Coefficient of x^{2} in a (1 - 2 x)^{18} \\ + Coefficient of x in b (1 - 2 x)^{18} \\ =^{18} C_{3} \cdot 2^{3} + a^{18} C_{2} \cdot 2^{2} - b^{18} C_{1} \cdot 2 \end{matrix}$

Given, coefficient of $x^{3} = 0$

$\Rightarrow^{18} C_{3} \cdot 2^{3} + a^{18} C_{2} \cdot 2^{2} - b^{18} C_{1} \cdot 2 = 0$

$\Rightarrow - \frac{18 \times 17 \times 16}{3 \times 2} \cdot 8 + a \cdot \frac{18 \times 17}{2} \cdot 2^{2} - b \cdot 18 \cdot 2 = 0$

$\Rightarrow 17 a - b = \frac{34 \times 16}{3}$ …….(i)

Similarly, coefficient of $x^{4} = 0$

$\begin{aligned} \Rightarrow & ^{18} C_{4} \cdot 2^{4} - a \cdot^{18} C_{3} 2^{3} + b \cdot^{18} C_{2} \cdot 2^{2} = 0 \\ ∴ & 32 a - 3 b = 240 \dots \dots . (i i) \end{aligned}$

On solving Eqs. (i) and (ii), we get

$a = 16, b = \frac{272}{3}$

Binomial Theorem 1 Question 25

27.

Answer:

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Binomial Theorem 1 Question 25

27.

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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