| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Applied recurrence modeling |
| Difficulty | Challenging +1.2 This is a standard Further Maths recurrence relation question requiring proof by induction and monotonicity analysis. Part (i) follows a heavily guided approach ('by considering 3 - u_{n+1}'), making the algebraic manipulation straightforward. Part (ii) requires showing the sequence is increasing, which is routine once you compute u_{n+1} - u_n. While it involves multiple proof techniques and careful algebraic manipulation, the structure is typical for FP1 and the question provides strong scaffolding. Slightly above average difficulty due to the proof requirements and multi-part nature, but not requiring novel insight. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(u_1 < 3\) (given) | B1 | States base case |
| Assume that \(u_k < 3\) | B1 | States inductive hypothesis |
| \(3 - u_{k+1} = 3 - \frac{4u_k + 9}{u_k + 4} = \frac{-u_k + 3}{u_k + 4} > 0 \Rightarrow u_{k+1} < 3\) | M1 A1 | |
| Hence, by induction, \(u_n < 3\) for all \(n \geqslant 1\) | B1 | States conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(u_{n+1} - u_n = \frac{4u_n + 9}{u_n + 4} - u_n = \frac{-u_n^2 + 9}{u_n + 4}\) | M1 A1 | Considers \(u_{n+1} - u_n\) |
| So \(u_n < 3 \Rightarrow u_{n+1} - u_n > 0\) | B1 | Uses \(u_n < 3\) |
## Question 3:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_1 < 3$ (given) | B1 | States base case |
| Assume that $u_k < 3$ | B1 | States inductive hypothesis |
| $3 - u_{k+1} = 3 - \frac{4u_k + 9}{u_k + 4} = \frac{-u_k + 3}{u_k + 4} > 0 \Rightarrow u_{k+1} < 3$ | M1 A1 | |
| Hence, by induction, $u_n < 3$ for all $n \geqslant 1$ | B1 | States conclusion |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_{n+1} - u_n = \frac{4u_n + 9}{u_n + 4} - u_n = \frac{-u_n^2 + 9}{u_n + 4}$ | M1 A1 | Considers $u_{n+1} - u_n$ |
| So $u_n < 3 \Rightarrow u_{n+1} - u_n > 0$ | B1 | Uses $u_n < 3$ |
---3 The sequence of positive numbers $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is such that $u _ { 1 } < 3$ and, for $n \geqslant 1$,
$$u _ { n + 1 } = \frac { 4 u _ { n } + 9 } { u _ { n } + 4 }$$
(i) By considering $3 - u _ { n + 1 }$, or otherwise, prove by mathematical induction that $u _ { n } < 3$ for all positive integers $n$.\\
(ii) Show that $u _ { n + 1 } > u _ { n }$ for $n \geqslant 1$.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q3}}