1. (a) Find the binomial expansion of $$\frac{1}{\sqrt{9 - 10x}}$$ where $|x| < \frac{9}{10}$, in ascending powers of $x$ up to and including the term in $x^2$ - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 8

To find the binomial expansion of $\frac{1}{\sqrt{9 - 10x}}$, we first rewrite it:

$\frac{1}{\sqrt{9(1 - \frac{10}{9}x)}} = \frac{1}{3}\cdot (9(1 - \frac{10}{9}x))^{-\frac{1}{2}}$.

Using the binomial expansion for $(1 + u)^{n}$ where $n = -\frac{1}{2}$ and $u = -\frac{10}{9}x$ gives:

$(1 - \frac{10}{9}x)^{-\frac{1}{2}} = 1 + \frac{1}{2}\left(-\frac{10}{9}x\right) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}\left(-\frac{10}{9}x\right)^2 + ...$

Calculating the first three terms:

1. The constant term: 1
2. The coefficient of $x$: $-\frac{10}{18} = -\frac{5}{9}$
3. The coefficient of $x^2$: $\frac{5}{9} \cdot \frac{100}{81} = \frac{500}{729}$

Thus, the expansion to the second order term is given by:

$\frac{1}{\sqrt{9 - 10x}} = \frac{1}{3} \left(1 - \frac{5}{9}x + \frac{500}{1458}x^2 + ...\right)$

Therefore:

$\frac{1}{\sqrt{9 - 10x}} = \frac{1}{3} - \frac{5}{27}x + \frac{500}{4374}x^2 + ...$.