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






====Go To====
<text align="left"><btn type="primary">[[math-11-kpk:sol:unit07:ex7-3-p2 |< Question 2  ]]</btn></text>
<text align="right"><btn type="success">[[math-11-kpk:sol:unit07:ex7-3-p4|Question 4 >]]</btn></text>