| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Sum from n+1 to 2n or similar range |
| Difficulty | Challenging +1.8 Part (i) is a standard method of differences question requiring partial fractions and telescoping sum recognition. Part (ii) is significantly harder, requiring substitution of the result from (i), algebraic manipulation of sum limits, and careful limit evaluation with N² terms—this demands strong technical facility and insight beyond routine application. |
| Spec | 4.06b Method of differences: telescoping series8.01d Sequence limits: limit of nth term as n tends to infinity, steady-states |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{(3r+1)(3r-2)} = \frac{1}{3}\left(\frac{1}{3r-2} - \frac{1}{3r+1}\right)\) | M1 A1 | Finds partial fractions |
| \(\sum_{r=1}^{N} \frac{1}{(3r+1)(3r-2)} = \frac{1}{3}\left(\frac{1}{1} - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \cdots \frac{1}{3N-2} - \frac{1}{3N+1}\right)\) | M1 | At least 3 terms including final term |
| \(= \frac{1}{3}\left(1 - \frac{1}{3N+1}\right)\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=N+1}^{N^2} \frac{N}{(3r+1)(3r-2)} = \sum_{r=1}^{N^2} \frac{N}{(3r+1)(3r-2)} - \sum_{r=1}^{N} \frac{N}{(3r+1)(3r-2)}\) | M1 | Uses \(\sum_{r=N+1}^{N^2} = \sum_{r=1}^{N^2} - \sum_{r=1}^{N}\) |
| \(= \frac{N}{3} - \frac{N}{3(3N^2+1)} - \left(\frac{N}{3} - \frac{N}{3(3N+1)}\right)\) | M1 | Applies (i) |
| \(= \frac{N}{3(3N+1)} - \frac{N}{3(3N^2+1)} = \frac{N^3 - N^2}{(3N+1)(3N^2+1)}\) | A1 | Allow simplification to common denominator |
| \(\to \frac{1}{9}\) as \(N \to \infty\) | B1 |
## Question 4(i):
| $\frac{1}{(3r+1)(3r-2)} = \frac{1}{3}\left(\frac{1}{3r-2} - \frac{1}{3r+1}\right)$ | M1 A1 | Finds partial fractions |
| $\sum_{r=1}^{N} \frac{1}{(3r+1)(3r-2)} = \frac{1}{3}\left(\frac{1}{1} - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \cdots \frac{1}{3N-2} - \frac{1}{3N+1}\right)$ | M1 | At least 3 terms including final term |
| $= \frac{1}{3}\left(1 - \frac{1}{3N+1}\right)$ | A1 | AG |
## Question 4(ii):
| $\sum_{r=N+1}^{N^2} \frac{N}{(3r+1)(3r-2)} = \sum_{r=1}^{N^2} \frac{N}{(3r+1)(3r-2)} - \sum_{r=1}^{N} \frac{N}{(3r+1)(3r-2)}$ | M1 | Uses $\sum_{r=N+1}^{N^2} = \sum_{r=1}^{N^2} - \sum_{r=1}^{N}$ |
| $= \frac{N}{3} - \frac{N}{3(3N^2+1)} - \left(\frac{N}{3} - \frac{N}{3(3N+1)}\right)$ | M1 | Applies (i) |
| $= \frac{N}{3(3N+1)} - \frac{N}{3(3N^2+1)} = \frac{N^3 - N^2}{(3N+1)(3N^2+1)}$ | A1 | Allow simplification to common denominator |
| $\to \frac{1}{9}$ as $N \to \infty$ | B1 | |
---4 (i) Use the method of differences to show that $\sum _ { r = 1 } ^ { N } \frac { 1 } { ( 3 r + 1 ) ( 3 r - 2 ) } = \frac { 1 } { 3 } - \frac { 1 } { 3 ( 3 N + 1 ) }$.\\
(ii) Find the limit, as $N \rightarrow \infty$, of $\sum _ { r = N + 1 } ^ { N ^ { 2 } } \frac { N } { ( 3 r + 1 ) ( 3 r - 2 ) }$.\\
\hfill \mbox{\textit{CAIE FP1 2019 Q4 [8]}}