| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | November |
| 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 Further Maths question requiring geometric series summation with complex numbers and De Moivre's theorem. Part (a) is a straightforward application of the GP formula with algebraic manipulation. Part (b) requires using Euler's form and equating real parts, which is a standard technique in Further Maths. While it involves multiple steps and complex number manipulation, these are well-practiced techniques for Further Maths students, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 4.02q De Moivre's theorem: multiple angle formulae4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z^2 + z^2(z^2) + \ldots + z^2(z^2)^{n-1} = \frac{z^2(z^{2n}-1)}{z^2-1}\) | M1 | Uses sum of geometric series |
| \(\frac{z^2(z^{2n}-1)}{z^2-1} \times \frac{z^{-1}}{z^{-1}} = \frac{z^{2n+1}-z}{z-z^{-1}}\) | A1 | Divides numerator and denominator by \(z\). Must see at least \(\frac{z^{2n+2}-z^2}{z^2-1}\), AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z - z^{-1} = 2i\sin\theta\) | B1 | Simplifies denominator |
| \(\frac{z^{2n+1}-z}{z-z^{-1}} = \frac{\cos(2n+1)\theta + i\sin(2n+1)\theta - \cos\theta - i\sin\theta}{2i\sin\theta}\) | M1 A1 | Applies de Moivre's theorem to numerator |
| \(\sum_{r=1}^{n}\cos(2r\theta) = \frac{\sin(2n+1)\theta - \sin\theta}{2\sin\theta} = \frac{\sin(2n+1)\theta}{2\sin\theta} - \frac{1}{2}\) | M1 | Equates real parts |
| \(1 + 2\sum_{r=1}^{n}\cos(2r\theta) = \frac{\sin(2n+1)\theta}{\sin\theta}\) | A1 | AG |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z^2 + z^2(z^2) + \ldots + z^2(z^2)^{n-1} = \frac{z^2(z^{2n}-1)}{z^2-1}$ | M1 | Uses sum of geometric series |
| $\frac{z^2(z^{2n}-1)}{z^2-1} \times \frac{z^{-1}}{z^{-1}} = \frac{z^{2n+1}-z}{z-z^{-1}}$ | A1 | Divides numerator and denominator by $z$. Must see at least $\frac{z^{2n+2}-z^2}{z^2-1}$, AG |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z - z^{-1} = 2i\sin\theta$ | B1 | Simplifies denominator |
| $\frac{z^{2n+1}-z}{z-z^{-1}} = \frac{\cos(2n+1)\theta + i\sin(2n+1)\theta - \cos\theta - i\sin\theta}{2i\sin\theta}$ | M1 A1 | Applies de Moivre's theorem to numerator |
| $\sum_{r=1}^{n}\cos(2r\theta) = \frac{\sin(2n+1)\theta - \sin\theta}{2\sin\theta} = \frac{\sin(2n+1)\theta}{2\sin\theta} - \frac{1}{2}$ | M1 | Equates real parts |
| $1 + 2\sum_{r=1}^{n}\cos(2r\theta) = \frac{\sin(2n+1)\theta}{\sin\theta}$ | A1 | AG |7
\begin{enumerate}[label=(\alph*)]
\item Show that $\sum _ { r = 1 } ^ { n } z ^ { 2 r } = \frac { z ^ { 2 n + 1 } - z } { z - z ^ { - 1 } }$, for $z \neq 0,1 , - 1$.
\item By letting $z = \cos \theta + i \sin \theta$, show that, if $\sin \theta \neq 0$,
$$1 + 2 \sum _ { r = 1 } ^ { n } \cos ( 2 r \theta ) = \frac { \sin ( 2 n + 1 ) \theta } { \sin \theta }$$
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q7 [7]}}