5. (a) Expand $ \frac{1}{\sqrt{4-3x}} $, where $|x| < \frac{1}{3}$, in ascending powers of $x$ up to and including the term in $x^2$. Simplify each term. (b) Hence, or otherwise, find the first 3 terms in the expansion of $ \sqrt{x+8} $ as a series in ascending powers of $x$.

A-Level Edexcel Maths Pure Partial Fractions: 5. (a) Expand \( \frac{1}{\sqrt{

5. (a) Expand $ \frac{1}{\sqrt{4-3x}} $, where $|x| < \frac{1}{3}$, in ascending powers of $x$ up to and including the term in $x^2$ - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 7

Question 7

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5. (a) Expand $ \frac{1}{\sqrt{4-3x}} $, where $|x| < \frac{1}{3}$, in ascending powers of $x$ up to and including the term in $x^2$. Simplify each term. (b... show full transcript

Worked Solution & Example Answer:5. (a) Expand $ \frac{1}{\sqrt{4-3x}} $, where $|x| < \frac{1}{3}$, in ascending powers of $x$ up to and including the term in $x^2$ - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 7

Step 1

Expand $ \frac{1}{\sqrt{4-3x}} $ in ascending powers of $x$

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Answer

To expand ( \frac{1}{\sqrt{4-3x}} ), we first rewrite it as:

$\frac{1}{\sqrt{4}} \cdot \frac{1}{\sqrt{1-\frac{3x}{4}}} = \frac{1}{2} \cdot (1-\frac{3x}{4})^{-\frac{1}{2}}$

Now, we use the binomial expansion formula ((1 + u)^{n} \approx 1 + nu + \frac{n(n-1)}{2}u^2 + ...) where (u = -\frac{3x}{4}) and (n = -\frac{1}{2}):

$\Rightarrow \frac{1}{2} \left(1 + -\frac{1}{2}\left(-\frac{3x}{4}\right) + \frac{-\frac{1}{2}(-\frac{3}{4}-1)}{2!}\left(-\frac{3x}{4}\right)^2\right)$

Calculating the series:

First term: ( \frac{1}{2} )
Second term: ( -\frac{1}{2} \cdot -\frac{3x}{4} = \frac{3x}{8} )
Third term: ( \frac{-\frac{1}{2} \cdot -\frac{1}{4}}{2} \left(-\frac{3x}{4}\right)^2 = \frac{3^2x^2}{2\cdot4^2} = \frac{27x^2}{256} )

So, the expansion gives:

$\frac{1}{2} + \frac{3x}{8} + \frac{27x^2}{256}$

Step 2

Find the first 3 terms in the expansion of $ \sqrt{x+8} $

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Answer

To find the first three terms of ( \sqrt{x+8} ), we first rewrite it as ( \sqrt{8(1 + \frac{x}{8})} = 2\sqrt{2} \sqrt{1 + \frac{x}{8}} ).

Using the binomial expansion for ( \sqrt{1+u} ) where ( u = \frac{x}{8} ):

$\sqrt{1 + u} \approx 1 + \frac{1}{2}u - \frac{1}{8}u^2$

Substituting ( u ):

$\sqrt{1 + \frac{x}{8}} \approx 1 + \frac{1}{2}\left(\frac{x}{8}\right) - \frac{1}{8}\left(\frac{x}{8}\right)^2 = 1 + \frac{x}{16} - \frac{x^2}{512}$ .

Thus:

$\sqrt{x+8} \approx 2\sqrt{2}\left(1 + \frac{x}{16} - \frac{x^2}{512}\right) = 2\sqrt{2} + \frac{2\sqrt{2}}{16}x - \frac{2\sqrt{2}}{512}x^2$ .

Therefore, the first three terms are:

$2\sqrt{2} + \frac{\sqrt{2}}{8}x - \frac{\sqrt{2}}{256}x^2$ .

5. (a) Expand \( \frac{1}{\sqrt{4-3x}} \), where \(|x| < \frac{1}{3}\), in ascending powers of \(x\) up to and including the term in \(x^2\) - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 7

Worked Solution & Example Answer:5. (a) Expand \( \frac{1}{\sqrt{4-3x}} \), where \(|x| < \frac{1}{3}\), in ascending powers of \(x\) up to and including the term in \(x^2\) - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 7

Expand \( \frac{1}{\sqrt{4-3x}} \) in ascending powers of \(x\)

Find the first 3 terms in the expansion of \( \sqrt{x+8} \)

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