| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Sigma Notation Manipulation |
| Difficulty | Standard +0.8 This question requires understanding that u_r = S_r - S_{r-1} and manipulating sigma notation algebraically. Part (i) involves substituting 2n and n into the formula and subtracting, requiring careful algebraic expansion. Part (ii) requires the insight to find the general term from the sum formula, which is a non-routine technique for A-level students. While the algebra itself is manageable, the conceptual approach and multi-step manipulation elevate this above average difficulty. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{n+1}^{2n} u_r = (2n)^2(4n+3) - n^2(2n+3)\) | M1 | Method mark for using \(S_{2n} - S_n\) |
| \(= 14n^3 + 9n^2\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(u_r = r^2(2r+3) - (r-1)^2(2r+1)\) | M1A1 | Method mark for using \(S_r - S_{r-1}\) OE |
| \(= 6r^2 - 1\) | A1 | SR: CAO B1 without wrong working |
| Total: 3 |
## Question 1:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{n+1}^{2n} u_r = (2n)^2(4n+3) - n^2(2n+3)$ | M1 | Method mark for using $S_{2n} - S_n$ |
| $= 14n^3 + 9n^2$ | A1 | |
| **Total: 2** | | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_r = r^2(2r+3) - (r-1)^2(2r+1)$ | M1A1 | Method mark for using $S_r - S_{r-1}$ OE |
| $= 6r^2 - 1$ | A1 | SR: CAO B1 without wrong working |
| **Total: 3** | | |
---1 It is given that $\sum _ { r = 1 } ^ { n } u _ { r } = n ^ { 2 } ( 2 n + 3 )$, where $n$ is a positive integer.\\
(i) Find $\sum _ { r = n + 1 } ^ { 2 n } u _ { r }$.\\
(ii) Find $u _ { r }$.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q1 [5]}}