| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Standard +0.3 This is a standard method of differences question where the partial fraction decomposition is given in part (i), requiring only algebraic verification. Part (ii) involves telescoping the series, which is a routine technique once the identity is established. The algebra is straightforward and the method is a textbook application, making this slightly easier than average for A-level Further Maths. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| M1 | Use correct denominator or partial fractions | |
| A1 | Obtain given answer convincingly | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| M1 | Express at least \(1^{st}\) two and last term using (i) | |
| A1 | All terms correct | |
| M1 | Show correct terms cancelling | |
| \(\frac{1}{2} - \frac{1}{6n+2}\) | A1 | Obtain correct unsimplified answer |
| M1 | Include \(\frac{1}{3}\) and combine their sum as a single fraction | |
| A1 | Obtain given answer | |
| [6] |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | Use correct denominator or partial fractions |
| | A1 | Obtain **given** answer convincingly |
| **[2]** | | |
---
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | Express at least $1^{st}$ two and last term using (i) |
| | A1 | All terms correct |
| | M1 | Show correct terms cancelling |
| $\frac{1}{2} - \frac{1}{6n+2}$ | A1 | Obtain correct unsimplified answer |
| | M1 | Include $\frac{1}{3}$ and combine their sum as a single fraction |
| | A1 | Obtain **given** answer |
| **[6]** | | |
---9 (i) Show that $\frac { 1 } { 3 r - 1 } - \frac { 1 } { 3 r + 2 } \equiv \frac { 3 } { ( 3 r - 1 ) ( 3 r + 2 ) }$.\\
(ii) Hence show that $\sum _ { r = 1 } ^ { 2 n } \frac { 1 } { ( 3 r - 1 ) ( 3 r + 2 ) } = \frac { n } { 2 ( 3 n + 1 ) }$.
\hfill \mbox{\textit{OCR FP1 2013 Q9 [8]}}