Use the binomial expansion to show that $$\sqrt{\frac{1+x}{1-x}} = 1 + x + \frac{1}{2}x^2, \ |x| < 1$$ (b) Substitute $x = \frac{1}{26}$ into $$\sqrt{\frac{1+x}{1-x}} = 1 + x + \frac{1}{2}x^2$$ to obtain an approximation to $\sqrt{3}$ Give your answer in the form $\frac{a}{b}$ where $a$ and $b$ are integers. - Edexcel - A-Level Maths Pure - Question 3 - 2013 - Paper 9

To show that $\sqrt{\frac{1+x}{1-x}} = \frac{(1+x)^{1/2}}{(1-x)^{1/2}}$ we can apply the binomial expansion on both the numerator and the denominator.

1. Expand the numerator using the binomial expansion:

   • For $(1+x)^{1/2}$: $(1+x)^{1/2} \approx 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \cdots$

2. Expand the denominator using the binomial expansion:

   • For $(1-x)^{1/2}$: $(1-x)^{1/2} \approx 1 - \frac{1}{2}x - \frac{1}{8}x^2 + \cdots$

3. Combine the expansions:

   • The expression becomes: $\sqrt{\frac{1+x}{1-x}} \approx \frac{1 + \frac{1}{2}x - \frac{1}{8}x^2}{1 - \frac{1}{2}x - \frac{1}{8}x^2}$

4. Perform polynomial long division or simplify to get: $\approx 1 + x + \frac{1}{2}x^2$

Hence, we have shown that: $\sqrt{\frac{1+x}{1-x}} = 1 + x + \frac{1}{2}x^2, \ |x| < 1$