| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Sum geometric series with complex terms |
| Difficulty | Challenging +1.8 This is a sophisticated multi-part Further Maths question requiring geometric series with complex terms, De Moivre's theorem manipulation, and extraction of imaginary parts. Part (b) requires proof using polar form, part (c) demands equating imaginary parts after applying the geometric series formula, and part (d) requires recognizing the specific substitution θ=π/3. While the individual techniques are standard for Further Maths, the extended chain of reasoning and the non-obvious connection between parts makes this significantly harder than average A-level questions but still within the expected range for Further Pure topics. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{w^n - 1}{w - 1}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1 + i\tan\theta)^k = \sec^k\theta(\cos\theta + i\sin\theta)^k = \sec^k\theta(\cos k\theta + i\sin k\theta)\) | M1 A1 | Applies de Moivre's theorem, AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{k=0}^{n-1}(1+i\tan\theta)^k = \dfrac{(1+i\tan\theta)^n - 1}{i\tan\theta}\) | M1 A1 | Applies part (a) |
| \(-\cot\theta\sec^n\theta(i\cos n\theta - \sin n\theta) + i\cot\theta\) | A1 | |
| \(\sum_{k=0}^{n-1}\sec^k\theta\sin k\theta = \cot\theta(1 - \sec^n\theta\cos n\theta)\) | M1 A1 | Takes imaginary part, AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\theta = \frac{1}{3}\pi\): \(\sum_{k=0}^{6m-1} 2^k\sin\left(\frac{1}{3}k\pi\right) = \frac{1}{3}\sqrt{3}\left(1 - 2^{6m}\cos(2m\pi)\right) = \frac{1}{3}\sqrt{3}(1 - 2^{6m})\) | M1 A1 | Sets \(\theta = \frac{1}{3}\pi\) |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{w^n - 1}{w - 1}$ | B1 | |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1 + i\tan\theta)^k = \sec^k\theta(\cos\theta + i\sin\theta)^k = \sec^k\theta(\cos k\theta + i\sin k\theta)$ | M1 A1 | Applies de Moivre's theorem, AG |
---
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{k=0}^{n-1}(1+i\tan\theta)^k = \dfrac{(1+i\tan\theta)^n - 1}{i\tan\theta}$ | M1 A1 | Applies part (a) |
| $-\cot\theta\sec^n\theta(i\cos n\theta - \sin n\theta) + i\cot\theta$ | A1 | |
| $\sum_{k=0}^{n-1}\sec^k\theta\sin k\theta = \cot\theta(1 - \sec^n\theta\cos n\theta)$ | M1 A1 | Takes imaginary part, AG |
---
## Question 7(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\theta = \frac{1}{3}\pi$: $\sum_{k=0}^{6m-1} 2^k\sin\left(\frac{1}{3}k\pi\right) = \frac{1}{3}\sqrt{3}\left(1 - 2^{6m}\cos(2m\pi)\right) = \frac{1}{3}\sqrt{3}(1 - 2^{6m})$ | M1 A1 | Sets $\theta = \frac{1}{3}\pi$ |7
\begin{enumerate}[label=(\alph*)]
\item State the sum of the series $1 + \mathrm { w } + \mathrm { w } ^ { 2 } + \mathrm { w } ^ { 3 } + \ldots + \mathrm { w } ^ { \mathrm { n } - 1 }$, for $w \neq 1$.
\item Show that $( 1 + i \tan \theta ) ^ { k } = \sec ^ { k } \theta ( \cos k \theta + i \sin k \theta )$, where $\theta$ is not an integer multiple of $\frac { 1 } { 2 } \pi$.
\item By considering $\sum _ { \mathrm { k } = 0 } ^ { \mathrm { n } - 1 } ( 1 + \mathrm { i } \tan \theta ) ^ { \mathrm { k } }$, show that
$$\sum _ { k = 0 } ^ { n - 1 } \sec ^ { k } \theta \sin k \theta = \cot \theta \left( 1 - \sec ^ { n } \theta \cos n \theta \right)$$
provided $\theta$ is not an integer multiple of $\frac { 1 } { 2 } \pi$.
\item Hence find $\sum _ { k = 0 } ^ { 6 m - 1 } 2 ^ { k } \sin \left( \frac { 1 } { 3 } k \pi \right)$ in terms of $m$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q7 [10]}}