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

To expand $\frac{1}{\sqrt{4-3x}}$, we can rewrite it as: $\frac{1}{\sqrt{4(1 - \frac{3x}{4})}} = \frac{1}{\sqrt{4}} \cdot \frac{1}{\sqrt{1 - \frac{3x}{4}}} = \frac{1}{2} \sum_{n=0}^{\infty} {\frac{1}{2} \choose n} \left( -\frac{3x}{4} \right)^{n}$

Using the binomial expansion, we calculate a few terms:

• For $n=0$: $1$
• For $n=1$: $-\frac{3x}{4} \cdot 1 = -\frac{3x}{4}$
• For $n=2$: $\frac{1}{2} \cdot \frac{1}{1!} \cdot \left(-\frac{3x}{4}\right)^{2} = \frac{9x^2}{32}$

So, we have: $\frac{1}{\sqrt{4-3x}} = \frac{1}{2} \left( 1 - \frac{3x}{4} + \frac{9x^2}{32} \right) = \frac{1}{2} - \frac{3x}{8} + \frac{9x^2}{64}$