| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Geometric Series with Complex Numbers |
| Difficulty | Challenging +1.2 This is a standard Further Maths question combining geometric series with De Moivre's theorem. Part (a) requires applying the GP formula and algebraic manipulation, while part (b) uses the standard technique of substituting Euler form and taking imaginary parts. Both parts follow well-established methods taught in Further Pure modules, though the algebraic manipulation requires care. Slightly above average difficulty due to the multi-step nature and Further Maths content, but this is a textbook application rather than requiring novel insight. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z^4 + z^4(z^4) + \ldots + z^4(z^4)^{n-1} = \frac{z^4(z^{4n}-1)}{z^4 - 1}\) | M1 | Uses sum of geometric series |
| \(\frac{z^4(z^{4n}-1)}{z^4-1} \times \frac{z^{-2}}{z^{-2}} = \frac{z^{4n+2} - z^2}{z^2 - z^{-2}}\) | A1 | Divides numerator and denominator by \(z^2\), AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z^2 - z^{-2} = 2i\sin 2\theta\) | B1 | Simplifies denominator |
| \(\frac{z^{4n+2} - z^2}{z^2 - z^{-2}} = \frac{\cos(4n+2)\theta + i\sin(4n+2)\theta - \cos 2\theta - i\sin 2\theta}{2i\sin 2\theta}\) | M1 A1 | Applies de Moivre's theorem to numerator (used at least once for M1); accept \(z^2 = \cos^2\theta + 2i\sin\theta\cos\theta - \sin^2\theta\) |
| \(\sum_{r=1}^n \sin(4r\theta) = \frac{\cos(4n+2)\theta - \cos 2\theta}{-2\sin 2\theta}\) | M1 | Equates imaginary parts |
| \(\sum_{r=1}^n \sin(4r\theta) = \frac{\cos 2\theta - \cos(4n+2)\theta}{2\sin 2\theta}\) | A1 | AG; must see evidence of eliminating \(i\) from denominator |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z^4 + z^4(z^4) + \ldots + z^4(z^4)^{n-1} = \frac{z^4(z^{4n}-1)}{z^4 - 1}$ | M1 | Uses sum of geometric series |
| $\frac{z^4(z^{4n}-1)}{z^4-1} \times \frac{z^{-2}}{z^{-2}} = \frac{z^{4n+2} - z^2}{z^2 - z^{-2}}$ | A1 | Divides numerator and denominator by $z^2$, AG |
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## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z^2 - z^{-2} = 2i\sin 2\theta$ | B1 | Simplifies denominator |
| $\frac{z^{4n+2} - z^2}{z^2 - z^{-2}} = \frac{\cos(4n+2)\theta + i\sin(4n+2)\theta - \cos 2\theta - i\sin 2\theta}{2i\sin 2\theta}$ | M1 A1 | Applies de Moivre's theorem to numerator (used at least once for M1); accept $z^2 = \cos^2\theta + 2i\sin\theta\cos\theta - \sin^2\theta$ |
| $\sum_{r=1}^n \sin(4r\theta) = \frac{\cos(4n+2)\theta - \cos 2\theta}{-2\sin 2\theta}$ | M1 | Equates imaginary parts |
| $\sum_{r=1}^n \sin(4r\theta) = \frac{\cos 2\theta - \cos(4n+2)\theta}{2\sin 2\theta}$ | A1 | AG; must see evidence of eliminating $i$ from denominator |6
\begin{enumerate}[label=(\alph*)]
\item Show that $\sum _ { r = 1 } ^ { n } z ^ { 4 r } = \frac { z ^ { 4 n + 2 } - z ^ { 2 } } { z ^ { 2 } - z ^ { - 2 } }$, for $z ^ { 2 } \neq z ^ { - 2 }$.
\item By letting $z = \cos \theta + \mathrm { i } \sin \theta$, show that, if $\sin 2 \theta \neq 0$,
$$\sum _ { r = 1 } ^ { n } \sin ( 4 r \theta ) = \frac { \cos 2 \theta - \cos ( 4 n + 2 ) \theta } { 2 \sin 2 \theta }$$
\includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-12_2718_35_143_2012}
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q6 [7]}}