7. (a) Use the binomial expansion, in ascending powers of $x$, to show that $$\sqrt{(4 - x)} = 2 - \frac{1}{4} x + kx^{2} + ...$$ where $k$ is a rational constant to be found - Edexcel - A-Level Maths Pure - Question 11 - 2017 - Paper 2

To use the binomial expansion for $\sqrt{(4 - x)}$, first factor out the 4:

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

Now use the binomial expansion formula:

$(1 + u)^{n} = 1 + nu + \frac{n(n-1)}{2!}u^{2} + ...$

Here, set $u = -\frac{x}{4}$ and $n = \frac{1}{2}$:

$\sqrt{(1 - \frac{x}{4})} = 1 - \frac{1}{2}\cdot \left(-\frac{x}{4}\right) + \frac{1/2 (1/2 - 1)}{2!}\left(-\frac{x}{4}\right)^{2} + ...$

This simplifies to:

$1 + \frac{1}{8} x - \frac{1}{128} x^{2} + ...$

Thus, substituting back gives:

$\sqrt{(4 - x)} = 2\left(1 + \frac{1}{8} x - \frac{1}{128} x^{2} + ...\right) = 2 - \frac{1}{4} x + kx^{2} + ...$

where $k = -\frac{1}{64}$.