C4 June 2009 Q1
1. \[\mathrm{f}(x) = \frac{1}{\sqrt{(4 + x)}}, \qquad |x| < 4\]
Find the binomial expansion of \(\mathrm{f}(x)\) in ascending powers of \(x\), up to and including the term in \(x^3\). Give each coefficient as a simplified fraction. (6)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = \dfrac{1}{\sqrt{(4 + x)}} = (4 + x)^{-\frac{1}{2}}\) | M1 |
| \(= (4)^{-\frac{1}{2}}(1 + \ldots)^{\cdots}\) \(\dfrac{1}{2}(1 + \ldots)^{\cdots}\) or \(\dfrac{1}{2\sqrt{(1 + \ldots)}}\) | B1 |
| \(= \ldots\left(1 + \left(-\tfrac{1}{2}\right)\left(\dfrac{x}{4}\right) + \dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(\dfrac{x}{4}\right)^2 + \dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}\left(\dfrac{x}{4}\right)^3 + \ldots\right)\) ft their \(\left(\dfrac{x}{4}\right)\) | M1 A1ft |
| \(= \dfrac{1}{2} - \dfrac{1}{16}x, + \dfrac{3}{256}x^2 - \dfrac{5}{2048}x^3 + \ldots\) | A1, A1 |
| (6) | |
| (6 marks) |
Alternative
| \(\mathrm{f}(x) = \dfrac{1}{\sqrt{(4 + x)}} = (4 + x)^{-\frac{1}{2}}\) | M1 |
| \(= \underline{4^{-\frac{1}{2}}} + \left(-\tfrac{1}{2}\right)4^{-\frac{3}{2}}x + \dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{1.2}4^{-\frac{5}{2}}x^2 + \dfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{1.2.3}4^{-\frac{7}{2}}x^3 + \ldots\) | B1 M1 A1 |
| \(= \dfrac{1}{2} - \dfrac{1}{16}x, + \dfrac{3}{256}x^2 - \dfrac{5}{2048}x^3 + \ldots\) | A1, A1 |
| (6) |