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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Binomial Theorem

NDA 2024 · Mathematics · Binomial Theorem

What is the coefficient of x^ 10 in the expansion of (1-x^2)^ 20...

What is the coefficient of x10intheexpansionofx^{10} in the expansion of\left(1-x^2\right)^{20}(2−x2−1x2)−5\left(2-x^2-\frac{1}{x^2}\right)^{-5} ?

Options

  1. A-1
  2. B1
  3. C10
  4. DCoefficient of x10x^{10} does not exist

Correct answer

A. -1

Step-by-step solution

(1−x2)20[−(1x2+x2−2)]−5\left(1-x^2\right)^{20}\left[-\left(\frac{1}{x^2}+x^2-2\right)\right]^{-5}=−(1−x2)20(x−1x)−10=−(1−x2)20(x2−1)−10x10=−(1−x2)10⋅x10Tr+1=−10Cr(1)10−r(−x2)10−r⋅x10=(−1)11−r⋅10Crx30−2r∵30−2r=10⇒r=10∴ Coefficient of  x10=(−1)110C10=−1×1−1. \begin{aligned} & =-\left(1-x^2\right)^{20}\left(x-\frac{1}{x}\right)^{-10} \\ & =-\left(1-x^2\right)^{20}\left(x^2-1\right)^{-10} x^{10} \\ & =-\left(1-x^2\right)^{10} \cdot x^{10} \\ & \mathrm{ T}_{r+1}=-{ }^{10} \mathrm{C}_r(1)^{10-r}\left(-x^2\right)^{10-r} \cdot x^{10} \\ & =(-1)^{11-r} \cdot{ }^{10} \mathrm{C}_r x^{30-2 r} \\ & \because \quad 30-2 r=10 \\ & \Rightarrow \quad r=10 \\ & \therefore \text{ Coefficient\text{ of } } x^{10}=(-1)^1{ }^{10} \mathrm{C}_{10}=-1 \times 1-1 \text{. } \end{aligned}$$
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