| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| 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 manipulation, partial fractions recognition, and understanding of series convergence. While the individual steps are methodical (verify identity, apply telescoping, evaluate limit), it requires insight to recognize the telescoping pattern and careful bookkeeping of terms. The infinite series evaluation adds conceptual depth beyond routine FP1 exercises, placing it moderately above average difficulty. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| B1 [1] | Obtain given answer correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| M1 | Express at least 1st two and last two terms using (i) | |
| A1 | 1st two terms correct | |
| A1 | Last two terms correct | |
| M1 | Show that correct terms cancel | |
| \(\frac{3}{2} - \frac{1}{n} - \frac{1}{(n+1)}\) | A1 [5] | Obtain correct answer, a.e.f. in terms of \(n\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| B1ft | Sum to infinity stated or implied or start at 1000 as in (ii) | |
| M1 | \(S_\infty\) − their (ii) with \(n = 999\) or \(1000\) or show correct cancelling | |
| \(\frac{1999}{999000}\) | A1 [3] | Obtain correct answer, a.e.f. (condone 0.002) |
## Question 7:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| | B1 **[1]** | Obtain **given** answer correctly |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | Express at least 1st two and last two terms using (i) |
| | A1 | 1st two terms correct |
| | A1 | Last two terms correct |
| | M1 | Show that correct terms cancel |
| $\frac{3}{2} - \frac{1}{n} - \frac{1}{(n+1)}$ | A1 **[5]** | Obtain correct answer, a.e.f. in terms of $n$ |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| | B1ft | Sum to infinity stated or implied or start at 1000 as in (ii) |
| | M1 | $S_\infty$ − their (ii) with $n = 999$ or $1000$ or show correct cancelling |
| $\frac{1999}{999000}$ | A1 **[3]** | Obtain correct answer, a.e.f. (condone 0.002) |
---7 (i) Show that $\frac { 1 } { r - 1 } - \frac { 1 } { r + 1 } \equiv \frac { 2 } { r ^ { 2 } - 1 }$.\\
(ii) Hence find an expression, in terms of $n$, for $\sum _ { r = 2 } ^ { n } \frac { 2 } { r ^ { 2 } - 1 }$.\\
(iii) Find the value of $\sum _ { r = 1000 } ^ { \infty } \frac { 2 } { r ^ { 2 } - 1 }$.
\hfill \mbox{\textit{OCR FP1 2011 Q7 [9]}}