| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Infinite series convergence and sum |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring multiple sophisticated techniques: deriving a cubic sum formula using standard results, applying method of differences to find a general term, and evaluating a limit involving the product of sequences. The final part requires algebraic manipulation of complex expressions and understanding of dominant terms in limits, going well beyond routine A-level work. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=1}^{N}(5r+1)(5r+6) = 25\sum_{r=1}^{N}r^2 + 35\sum_{r=1}^{N}r + 6N\) | M1 | Expands |
| \(25\left(\frac{1}{6}N(N+1)(2N+1)\right) + 35\left(\frac{1}{2}N(N+1)\right) + 6N\) | M1 | Substitutes formulae for \(\sum r\) and \(\sum r^2\) |
| \(= N\left(\frac{25}{6}(2N^2+3N+1)+\frac{35}{2}N+\frac{35}{2}+6\right) = \frac{1}{3}N(25N^2+90N+83)\) | A1 | Simplifies to given answer (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{(5r+1)(5r+6)} = \frac{1}{5}\left(\frac{1}{5r+1} - \frac{1}{5r+6}\right)\) | M1 A1 | Finds partial fractions |
| \(T_N = \frac{1}{5}\left(\frac{1}{6} - \frac{1}{11} + \frac{1}{11} - \frac{1}{16} + \cdots + \frac{1}{5N+1} - \frac{1}{5N+6}\right)\) | M1 | Expresses terms as differences |
| \(\frac{1}{5}\left(\frac{1}{6} - \frac{1}{5N+6}\right) = \frac{1}{30} - \frac{1}{5(5N+6)}\) | A1 | At least 3 terms including last |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{S_N}{N^3}T_N \to \frac{25}{3} \times \frac{1}{30} = \frac{5}{18}\) | M1 A1 | Divides \(S_N\) by \(N^3\) and takes limits as \(N\to\infty\) |
## Question 5:
### Part 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{N}(5r+1)(5r+6) = 25\sum_{r=1}^{N}r^2 + 35\sum_{r=1}^{N}r + 6N$ | M1 | Expands |
| $25\left(\frac{1}{6}N(N+1)(2N+1)\right) + 35\left(\frac{1}{2}N(N+1)\right) + 6N$ | M1 | Substitutes formulae for $\sum r$ and $\sum r^2$ |
| $= N\left(\frac{25}{6}(2N^2+3N+1)+\frac{35}{2}N+\frac{35}{2}+6\right) = \frac{1}{3}N(25N^2+90N+83)$ | A1 | Simplifies to given answer (AG) |
**Total: 3 marks**
### Part 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{(5r+1)(5r+6)} = \frac{1}{5}\left(\frac{1}{5r+1} - \frac{1}{5r+6}\right)$ | M1 A1 | Finds partial fractions |
| $T_N = \frac{1}{5}\left(\frac{1}{6} - \frac{1}{11} + \frac{1}{11} - \frac{1}{16} + \cdots + \frac{1}{5N+1} - \frac{1}{5N+6}\right)$ | M1 | Expresses terms as differences |
| $\frac{1}{5}\left(\frac{1}{6} - \frac{1}{5N+6}\right) = \frac{1}{30} - \frac{1}{5(5N+6)}$ | A1 | At least 3 terms including last |
**Total: 4 marks**
### Part 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{S_N}{N^3}T_N \to \frac{25}{3} \times \frac{1}{30} = \frac{5}{18}$ | M1 A1 | Divides $S_N$ by $N^3$ and takes limits as $N\to\infty$ |
**Total: 2 marks**
---(i) Use standard results from the List of Formulae (MF10) to show that
$$S _ { N } = \frac { 1 } { 3 } N \left( 25 N ^ { 2 } + 90 N + 83 \right)$$
(ii) Use the method of differences to express $T _ { N }$ in terms of $N$.\\
(iii) Find $\lim _ { N \rightarrow \infty } \left( N ^ { - 3 } S _ { N } T _ { N } \right)$.\\
\hfill \mbox{\textit{CAIE FP1 2019 Q5 [9]}}