| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Trigonometric method of differences |
| Difficulty | Challenging +1.2 This is a structured Further Maths question that guides students through a telescoping series proof. Part (i) requires applying given formulae (factor formulae for cos P ± cos Q) with careful algebraic manipulation, part (ii) is standard method of differences yielding a telescoping sum, and part (iii) requires recognizing that 1/cos(N) oscillates and doesn't tend to zero. While it involves multiple steps and Further Maths content, the question provides significant scaffolding and tests well-practiced techniques rather than requiring novel insight. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{4\sin\left(n-\frac{1}{2}\right)\sin\frac{1}{2}}{\cos(2n-1)+\cos 1} = \frac{2(\cos(n-1)-\cos n)}{2\cos n\cos(n-1)}\) | M1 | Uses formulae for \(\cos P \pm \cos Q\). |
| \(\frac{1}{\cos n} - \frac{1}{\cos(n-1)}\) | A1 | AG |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{n=1}^{N}\frac{4\sin\left(n-\frac{1}{2}\right)\sin\frac{1}{2}}{\cos(2n-1)+\cos 1} = \sum_{n=1}^{N}\frac{1}{\cos n}-\frac{1}{\cos(n-1)}\) \(= \frac{1}{\cos 1}-\frac{1}{1}+\frac{1}{\cos 2}-\frac{1}{\cos 1}+\ldots+\frac{1}{\cos N}-\frac{1}{\cos(N-1)}\) | M1 | Applies (i), shows enough terms and cancellation. |
| \(-1+\frac{1}{\cos N}\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos N\) oscillates as \(N\to\infty\) so \(u_1+u_2+u_3+\ldots\) does not converge. | B1 | States "oscillates" or refers to diverging values of \(\cos N\). |
| Total: 1 |
## Question 2:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{4\sin\left(n-\frac{1}{2}\right)\sin\frac{1}{2}}{\cos(2n-1)+\cos 1} = \frac{2(\cos(n-1)-\cos n)}{2\cos n\cos(n-1)}$ | M1 | Uses formulae for $\cos P \pm \cos Q$. |
| $\frac{1}{\cos n} - \frac{1}{\cos(n-1)}$ | A1 | AG |
| **Total: 2** | | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{n=1}^{N}\frac{4\sin\left(n-\frac{1}{2}\right)\sin\frac{1}{2}}{\cos(2n-1)+\cos 1} = \sum_{n=1}^{N}\frac{1}{\cos n}-\frac{1}{\cos(n-1)}$ $= \frac{1}{\cos 1}-\frac{1}{1}+\frac{1}{\cos 2}-\frac{1}{\cos 1}+\ldots+\frac{1}{\cos N}-\frac{1}{\cos(N-1)}$ | M1 | Applies (i), shows enough terms and cancellation. |
| $-1+\frac{1}{\cos N}$ | A1 | |
| **Total: 2** | | |
**Part (iii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos N$ oscillates as $N\to\infty$ so $u_1+u_2+u_3+\ldots$ does not converge. | B1 | States "oscillates" or refers to diverging values of $\cos N$. |
| **Total: 1** | | |2 Let $u _ { n } = \frac { 4 \sin \left( n - \frac { 1 } { 2 } \right) \sin \frac { 1 } { 2 } } { \cos ( 2 n - 1 ) + \cos 1 }$.\\
(i) Using the formulae for $\cos P \pm \cos Q$ given in the List of Formulae MF10, show that
$$u _ { n } = \frac { 1 } { \cos n } - \frac { 1 } { \cos ( n - 1 ) }$$
(ii) Use the method of differences to find $\sum _ { n = 1 } ^ { N } u _ { n }$.\\
(iii) Explain why the infinite series $u _ { 1 } + u _ { 2 } + u _ { 3 } + \ldots$ does not converge.\\
\hfill \mbox{\textit{CAIE FP1 2019 Q2 [5]}}