A-Level Edexcel Maths Pure Parametric Equations: 2. (a) Use the binomial expansio

2. (a) 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 $\sqrt{3}$ - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 9

Question 4

$2.-(a)-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-$\sqrt{3}$-Edexcel-A-Level Maths Pure-Question 4-2013-Paper 9.png$

2. (a) 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... show full transcript

Worked Solution & Example Answer:2. (a) 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 $\sqrt{3}$ - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 9

Step 1

Use the binomial expansion to show that $$\sqrt{\frac{1+x}{1-x}} = 1 + x + \frac{1}{2}x^2, \ |x| < 1$$

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Answer

To show that $\sqrt{\frac{1+x}{1-x}}$ can be expressed as $1 + x + \frac{1}{2}x^2$ , we start by rewriting the expression using the binomial theorem: $\sqrt{\frac{1+x}{1-x}} = \left(1+x\right)^{\frac{1}{2}} \left(1-x\right)^{-\frac{1}{2}}$ Applying the binomial expansion for both parts:

For $(1+x)^{\frac{1}{2}}$ :

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

$\approx 1 + \frac{1}{2}x + \frac{3}{8}x^2 + ...$

Multiplying these two expansions together, we focus on the first three terms:

$\left(1 + \frac{1}{2}x - \frac{1}{8}x^2\right) \left(1 + \frac{1}{2}x + \frac{3}{8}x^2\right)$

Calculating the product, we find:

$\approx 1 + x + \frac{1}{2} x^2$ valid for $|x| < 1$ .

Step 2

Substitute $x = \frac{1}{26}$ into $$\sqrt{\frac{1+x}{1-x}} = 1 + x + \frac{1}{2}x^2$$

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Answer

To find an approximation for $\sqrt{3}$ , we substitute $x = \frac{1}{26}$ into the expression:

Calculate:

$1 + \frac{1}{26} + \frac{1}{2}\left(\frac{1}{26}\right)^2$

Step-by-step:
- The first term is $1$ .
- The second term is: $\frac{1}{26} = \frac{1}{26}$
- The third term: $\frac{1}{2}\left(\frac{1}{26}\right)^2 = \frac{1}{2} \cdot \frac{1}{676} = \frac{1}{1352}$

Combining these results:

$1 + \frac{26}{676} + \frac{1}{1352} = \frac{1352}{1352} + \frac{26}{676}$

Converting the second term: $\frac{26}{676} = \frac{52}{1352}$

Thus, combining all: $\frac{1352 + 52}{1352} = \frac{1404}{1352}$

Therefore, the approximation to $\sqrt{3}$ is: $\sqrt{3} \approx \frac{1404}{1352}$ which can be simplified further.

Use the binomial expansion to show that $$\sqrt{\frac{1+x}{1-x}} = 1 + x + \frac{1}{2}x^2, \ |x| < 1$$

Substitute $x = \frac{1}{26}$ into $$\sqrt{\frac{1+x}{1-x}} = 1 + x + \frac{1}{2}x^2$$

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