| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Infinite series convergence and sum |
| Difficulty | Standard +0.8 This is a telescoping series question requiring algebraic verification, summation using differences, and infinite series evaluation. While the method is standard for FP1, it demands careful manipulation of partial fractions in reverse, recognition of the telescoping pattern, and precise handling of the limit. The multi-step nature and requirement to work with both finite and infinite sums places it moderately above average difficulty. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Combine with a correct denominator | M1 | |
| Obtain given answer correctly | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Express as differences using (i) | M1 | |
| Attempt this for at least first 3 terms | M1 | |
| First 3 terms all correct | A1 | |
| Last 2 terms all correct | A1 | |
| Show correct cancelling | M1 | |
| \(1 + \frac{1}{4} - \frac{1}{(n+1)^2} - \frac{1}{(n+2)^2}\) — obtain correct answer i.s.w. | A1 | Final answer must be in terms of \(n\) |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Start differences at \(n=5\) or \(S_\infty - S_4\) | M1 | \(1 + \frac{1}{4} - (1 - \frac{1}{4} - \frac{1}{5^2} - \frac{1}{6^2})\) |
| \(\frac{61}{900}\) — obtain correct answer with no errors seen | A1 | |
| [2] |
# Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Combine with a correct denominator | M1 | |
| Obtain **given** answer correctly | A1 | |
| **[2]** | | |
---
# Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Express as differences using (i) | M1 | |
| Attempt this for at least first 3 terms | M1 | |
| First 3 terms all correct | A1 | |
| Last 2 terms all correct | A1 | |
| Show correct cancelling | M1 | |
| $1 + \frac{1}{4} - \frac{1}{(n+1)^2} - \frac{1}{(n+2)^2}$ — obtain correct answer i.s.w. | A1 | Final answer must be in terms of $n$ |
| **[6]** | | |
---
# Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Start differences at $n=5$ or $S_\infty - S_4$ | M1 | $1 + \frac{1}{4} - (1 - \frac{1}{4} - \frac{1}{5^2} - \frac{1}{6^2})$ |
| $\frac{61}{900}$ — obtain correct answer **with no errors seen** | A1 | |
| **[2]** | | |
---6 (i) Show that $\frac { 1 } { r ^ { 2 } } - \frac { 1 } { ( r + 2 ) ^ { 2 } } \equiv \frac { 4 ( r + 1 ) } { r ^ { 2 } ( r + 2 ) ^ { 2 } }$.\\
(ii) Hence find an expression, in terms of $n$, for $\sum _ { r = 1 } ^ { n } \frac { 4 ( r + 1 ) } { r ^ { 2 } ( r + 2 ) ^ { 2 } }$.\\
(iii) Find $\sum _ { r = 5 } ^ { \infty } \frac { 4 ( r + 1 ) } { r ^ { 2 } ( r + 2 ) ^ { 2 } }$, giving your answer in the form $\frac { p } { q }$ where $p$ and $q$ are integers.
\hfill \mbox{\textit{OCR FP1 2014 Q6 [10]}}