| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 6 |
| 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 where the partial fractions decomposition is given. Part (i) requires recognizing the telescoping pattern and computing the sum (routine FP1 technique), while part (ii) is immediate once (i) is done. The question is slightly easier than average A-level difficulty since the hardest step (partial fractions) is provided, leaving only mechanical application of telescoping series. |
| Spec | 1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=1}^{100} \frac{1}{(5+3r)(2+3r)} = k\sum_{r=1}^{100}\left[\frac{1}{2+3r} - \frac{1}{5+3r}\right]\) | M1 | |
| \(= k\left[\left(\frac{1}{5} - \frac{1}{8}\right) + \left(\frac{1}{8} - \frac{1}{11}\right) + \ldots + \left(\frac{1}{302} - \frac{1}{305}\right)\right]\) | M1 | Write out terms (at least first and last terms in full) |
| A1 | ||
| \(= k\left(\frac{1}{5} - \frac{1}{305}\right)\) | M1 | Cancelling inner terms |
| \(= \frac{20}{305} = \frac{4}{61}\), oe | A1 | cao |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{15}\) | B1 | |
| [1] |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{100} \frac{1}{(5+3r)(2+3r)} = k\sum_{r=1}^{100}\left[\frac{1}{2+3r} - \frac{1}{5+3r}\right]$ | M1 | |
| $= k\left[\left(\frac{1}{5} - \frac{1}{8}\right) + \left(\frac{1}{8} - \frac{1}{11}\right) + \ldots + \left(\frac{1}{302} - \frac{1}{305}\right)\right]$ | M1 | Write out terms (at least first and last terms in full) |
| | A1 | |
| $= k\left(\frac{1}{5} - \frac{1}{305}\right)$ | M1 | Cancelling inner terms |
| $= \frac{20}{305} = \frac{4}{61}$, oe | A1 | cao |
| | **[5]** | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{15}$ | B1 | |
| | **[1]** | |
---5 You are given that $\frac { 3 } { ( 5 + 3 x ) ( 2 + 3 x ) } \equiv \frac { 1 } { 2 + 3 x } - \frac { 1 } { 5 + 3 x }$.\\
(i) Use this result to find $\sum _ { r = 1 } ^ { 100 } \frac { 1 } { ( 5 + 3 r ) ( 2 + 3 r ) }$, giving your answer as an exact fraction.\\
(ii) Write down the limit to which $\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 5 + 3 r ) ( 2 + 3 r ) }$ converges as $n$ tends to infinity.
\hfill \mbox{\textit{OCR MEI FP1 2013 Q5 [6]}}