SATHEE: Binomial Theorem 1 Question 1

Binomial Theorem 1 Question 1

1.

The coefficient of $x^{18}$ in the product $(1 + x) (1 - x)^{10} {(1 + x + x^{2})}^{9}$ is

(2019 Main, 12 April I)

(a) 84

(b) -126

(d) 126

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Answer:

Correct Answer: 1. (a)

Solution:

Given expression is

$\begin{aligned} (1 + x) & (1 - x)^{10} {(1 + x + x^{2})}^{9} \\ = & (1 + x) (1 - x) {[(1 - x) (1 + x + x^{2})]}^{9} \\ = & (1 - x^{2}) {(1 - x^{3})}^{9} \end{aligned}$

Now, coefficient of $x^{18}$ in the product

$\begin{aligned} (1 + x) (1 - x)^{10} {(1 + x + x^{2})}^{9} \\ = coefficient of x^{18} in the product (1 - x^{2}) {(1 - x^{3})}^{9} \\ = coefficient of x^{18} in {(1 - x^{3})}^{9} - coefficient of x^{16} in {(1 - x^{3})}^{9} \end{aligned}$

Since, $(r + 1)^{th}$ term in the expansion of

${(1 - x^{3})}^{9} is^{9} C_{r} {(- x^{3})}^{r} =^{9} C_{r} (- 1)^{r} x^{3 r}$

Now, for $x^{18}, 3 r = 18 \Rightarrow r = 6$

and for $x^{16}, 3 r = 16$

$\Rightarrow r = \frac{16}{3} \notin N$ .

$∴$ Required coefficient is $^{9} C_{6} = \frac{9!}{6! 3!} = \frac{9 \times 8 \times 7}{3 \times 2} = 84$

Binomial Theorem 1 Question 1

1.

Answer:

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Binomial Theorem 1 Question 1

1.

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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