Question 3 Exercise 7.3

Solutions of Question 3 of Exercise 7.3 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Q3 Expand $\sqrt{\frac{1-x}{1+x}}$ up-to $x^3$. Solution: Given $\sqrt{\frac{1-x}{1+x}}$ $$=(1-x)^{\frac{1}{2}}(1+x)^{-\frac{1}{2}} \text {. }$$

Applying binomial theorem, $$\begin{aligned} & (1-x)^{\frac{1}{2}}(1+x)^{\frac{1}{2}} \\ & =\left[1-\frac{x}{2}+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2 !}(-x)^2+\right. \\ & \left.\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3 !}(-x)^3+\ldots\right] \times \\ & {\left[1-\frac{x}{2}+\frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2 !} x^2+\right.} \\ & \left.\frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3 !} x^3+\ldots\right] \\ & =\left[1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16}+\ldots\right] \times \\ & {\left[1-\frac{x}{2}+\frac{3 x^2}{8}-\frac{5 x^3}{16}+\ldots\right]} \\ & \end{aligned}$$

Multiplying and writing terms up to $x^3$ $$-1 \quad \frac{x}{2}+\frac{3 x^2}{8} \quad \frac{5 x^3}{16}-\frac{x}{2}+\frac{x^2}{4}-$$ $$\frac{3 x^3}{16}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{x^3}{16}$$ $$\begin{aligned} & =1-\left(\frac{x}{2}+\frac{x}{2}\right)+\left(\frac{3+2-1}{8}\right) x^2 \\ & -\frac{8}{16} x^3 \\ & =1-x+\frac{x^2}{2}-\frac{x^3}{2} . \end{aligned}$$

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