A2 June 2024 Paper 2 Q18

AQACurrent spec4 marksDifferentiation & Maclaurin

18 In this question you may use results from the formulae booklet without proof.

Use the binomial series for \((1 + x)^n\) and the Maclaurin’s series for \(\sin x\) to find the series expansion for \(\dfrac{1}{(1 + \sin\theta)^4}\) up to and including the term in \(\theta^3\) [4 marks]

Mark scheme

Scheme	Marks	AO
Substitutes \(\sin\theta\) for \(x\) in the first four terms of the correct binomial series with \(n = -4\). Condone sign errors.	M1	1.1a
Substitutes in their binomial series the first two terms of the sine series in the first bracket, and the first term of the sine series in subsequent brackets.	M1	3.1a
Obtains the correct series. ISW Condone \(x\) used for \(\theta\)	A1	1.1b
Discards higher powers of \(\theta\) in a correct, fully simplified, series	A1	2.2a
	(4 marks)

Typical solution

\[\begin{aligned} (1 + \sin\theta)^{-4} &= 1 + (-4)\sin\theta + \frac{(-4)(-5)}{2}\sin^2\theta + \frac{(-4)(-5)(-6)}{6}\sin^3\theta + \ldots \\ &= 1 - 4\left(\theta - \frac{\theta^3}{6} + \ldots\right) + 10\left(\theta - \frac{\theta^3}{6} + \ldots\right)^2 \\ &\quad - 20\left(\theta - \frac{\theta^3}{6} + \ldots\right)^3 + \ldots \\ &= 1 - 4\theta + \frac{2\theta^3}{3} + 10\theta^2 - 20\theta^3 + \mathrm{o}(\theta^4) \end{aligned}\]\[\frac{1}{(1 + \sin\theta)^4} = 1 - 4\theta + 10\theta^2 - \frac{58}{3}\theta^3 + \ldots\]