7. (a) Find the first four terms, in ascending powers of x, of the binomial expansion of \[ \sqrt{4 - 9x} = 2 \sqrt{1 - \frac{9}{4}x} \] writing each term in simplest form - Edexcel - A-Level Maths Pure - Question 8 - 2022 - Paper 2

To find the first four terms of the binomial expansion of ( \sqrt{4 - 9x} ), we start by factoring out 4:

[ \sqrt{4 - 9x} = 2 \sqrt{1 - \frac{9}{4}x} ]

Now, we can use the binomial series for ( (1 + u)^n ), which gives:

[ (1 + u)^n \approx 1 + nu + \frac{n(n-1)}{2}u^2 + \frac{n(n-1)(n-2)}{6}u^3 + \ldots ]

Here, ( n = \frac{1}{2} ) and ( u = -\frac{9}{4}x ). Thus, we can write:

[ \sqrt{4 - 9x} = 2 \left( 1 - \frac{9}{8}x - \frac{27}{512}x^2 - \frac{729}{61440}x^3 + \ldots \right) ]

So the first four terms are:

1. [ 2 \cdot 1 = 2 ]
2. [ -\frac{9}{4} \cdot \frac{1}{2} \cdot 2 = -\frac{9}{8}x ]
3. [ \frac{1}{2} \cdot -\frac{9}{4} \cdot -\frac{9}{4} \cdot 2 = -\frac{27}{64}x^2 ]
4. [ \frac{1}{2} \cdot -\frac{9}{4} \cdot -\frac{9}{4} \cdot -\frac{27}{64} \cdot 6x^3 = -\frac{729}{512}x^3 ]

Thus, the first four terms in simplest form are: [ 2 - \frac{9}{8}x - \frac{27}{64}x^2 - \frac{729}{512}x^3 ]