| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| 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 +0.3 This is a standard telescoping series question with clear scaffolding. Part (i) is routine algebraic verification, part (ii) applies the standard telescoping technique to find a finite sum, and part (iii) uses the limit as nāā. While it requires understanding of partial fractions and series convergence, the method is well-practiced in FP1 and the question structure guides students through each step explicitly. |
| Spec | 4.04g Vector product: a x b perpendicular vector4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{2r+3-(2r+1)}{(2r+1)(2r+3)}\) or \(\frac{2r+3-2r-1}{(2r+1)(2r+3)} = \frac{2}{(2r+1)(2r+3)}\) | B1 | Establish given result correctly; might be a "\(0=0\)" type verification, or partial fractions |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{3}-\frac{1}{5}\) and \(\frac{1}{5}-\frac{1}{7}\) | M1 | Express terms as differences using (i) |
| M1 | Attempt this for at least 1st two and last term | |
| A1 | First two terms correct, do not penalise missing factor of 2 | |
| \(\frac{1}{2(n+1)}-\frac{1}{2(n+3)}\) | A1 | Last term correct, do not penalise missing factor of 2; allow in terms of e.g. \(r\) or \(k\) |
| M1 | Show correct cancelling | |
| \(\frac{n}{3(2n+3)}\) | A1 | Obtain correct single fraction (denominator could be expanded); must be in terms of \(n\). N.B. Be on look out for \(f(r)-f(r+1)\) approach |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| *Either:* \(\frac{1}{6}-\frac{n-1}{3(2n+1)}\) | M1 | Use sum to \(\infty\) minus sum to \(n-1\) or \(n\) |
| A1 | Obtain correct unsimplified expression, do not penalise missing factor of 2 | |
| \(\frac{1}{2(2n+1)}\) | A1 | Obtain correct single fraction (denominator could be expanded) |
| [3] | ||
| *Or:* \(\frac{1}{(2n+1)}\) | M1 | Start differences at \(r=n\) |
| A1 | Obtain correct remaining term | |
| \(\frac{1}{2(2n+1)}\) | A1 | Obtain correct single fraction (denominator could be expanded) |
## Question 8:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{2r+3-(2r+1)}{(2r+1)(2r+3)}$ or $\frac{2r+3-2r-1}{(2r+1)(2r+3)} = \frac{2}{(2r+1)(2r+3)}$ | B1 | Establish given result correctly; might be a "$0=0$" type verification, or partial fractions |
| **[1]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{3}-\frac{1}{5}$ and $\frac{1}{5}-\frac{1}{7}$ | M1 | Express terms as differences using (i) |
| | M1 | Attempt this for at least 1st two and last term |
| | A1 | First two terms correct, do not penalise missing factor of 2 |
| $\frac{1}{2(n+1)}-\frac{1}{2(n+3)}$ | A1 | Last term correct, do not penalise missing factor of 2; allow in terms of e.g. $r$ or $k$ |
| | M1 | Show correct cancelling |
| $\frac{n}{3(2n+3)}$ | A1 | Obtain correct single fraction (denominator could be expanded); must be in terms of $n$. N.B. Be on look out for $f(r)-f(r+1)$ approach |
| **[6]** | | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| *Either:* $\frac{1}{6}-\frac{n-1}{3(2n+1)}$ | M1 | Use sum to $\infty$ minus sum to $n-1$ or $n$ |
| | A1 | Obtain correct unsimplified expression, do not penalise missing factor of 2 |
| $\frac{1}{2(2n+1)}$ | A1 | Obtain correct single fraction (denominator could be expanded) |
| **[3]** | | |
| *Or:* $\frac{1}{(2n+1)}$ | M1 | Start differences at $r=n$ |
| | A1 | Obtain correct remaining term |
| $\frac{1}{2(2n+1)}$ | A1 | Obtain correct single fraction (denominator could be expanded) |(i) Show that $\frac { 1 } { 2 r + 1 } - \frac { 1 } { 2 r + 3 } \equiv \frac { 2 } { ( 2 r + 1 ) ( 2 r + 3 ) }$.\\
(ii) Hence find $\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) }$, giving your answer as a single fraction.\\
(iii) Find $\sum _ { r = n } ^ { \infty } \frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) }$, giving your answer as a single fraction.
\hfill \mbox{\textit{OCR FP1 2016 Q8 [10]}}