| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove inequality: recurrence sequence |
| Difficulty | Challenging +1.2 This is a standard Further Maths induction proof with a recursive sequence. Part (a) follows a guided approach ('by considering u_{n+1} - 4') making the algebraic manipulation straightforward. Part (b) requires showing the sequence is decreasing, which is routine once part (a) is established. The algebra is manageable and the structure is typical of Further Maths induction questions, placing it moderately above average difficulty. |
| Spec | 1.04e Sequences: nth term and recurrence relations4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(u_1 > 4\) (given) | B1 | States base case |
| Assume that \(u_k > 4\) for some positive integer \(k\) | B1 | States inductive hypothesis |
| Then \(u_{k+1} - 4 = \dfrac{u_k^2 + u_k + 12}{2u_k} - 4 = \dfrac{u_k^2 - 7u_k + 12}{2u_k}\) | M1 | Considers \(u_{k+1} - 4\), puts over common denominator |
| \(u_{k+1} - 4 = \dfrac{(u_k - 3)(u_k - 4)}{2u_k} > 0\) | A1 | |
| Hence, by induction, \(u_n > 4\) for all positive integers \(n\) | A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(u_{n+1} - u_n = \dfrac{-u_n^2 + u_n + 12}{2u_n} = \dfrac{-(u_n - 4)(u_n + 3)}{2u_n}\) | M1 A1 | Considers \(u_{n+1} - u_n\) |
| So \(u_n > 4 \Rightarrow u_{n+1} - u_n < 0\) | A1 | Uses \(u_n > 4\) |
| Total: 3 |
## Question 3:
### Part 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_1 > 4$ (given) | B1 | States base case |
| Assume that $u_k > 4$ for some positive integer $k$ | B1 | States inductive hypothesis |
| Then $u_{k+1} - 4 = \dfrac{u_k^2 + u_k + 12}{2u_k} - 4 = \dfrac{u_k^2 - 7u_k + 12}{2u_k}$ | M1 | Considers $u_{k+1} - 4$, puts over common denominator |
| $u_{k+1} - 4 = \dfrac{(u_k - 3)(u_k - 4)}{2u_k} > 0$ | A1 | |
| Hence, by induction, $u_n > 4$ for all positive integers $n$ | A1 | |
| **Total: 5** | | |
---
### Part 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_{n+1} - u_n = \dfrac{-u_n^2 + u_n + 12}{2u_n} = \dfrac{-(u_n - 4)(u_n + 3)}{2u_n}$ | M1 A1 | Considers $u_{n+1} - u_n$ |
| So $u_n > 4 \Rightarrow u_{n+1} - u_n < 0$ | A1 | Uses $u_n > 4$ |
| **Total: 3** | | |
---3 The sequence of positive numbers $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is such that $u _ { 1 } > 4$ and, for $n \geqslant 1$,
$$u _ { n + 1 } = \frac { u _ { n } ^ { 2 } + u _ { n } + 12 } { 2 u _ { n } }$$
\begin{enumerate}[label=(\alph*)]
\item By considering $\mathrm { u } _ { \mathrm { n } + 1 } - 4$, or otherwise, prove by mathematical induction that $\mathrm { u } _ { \mathrm { n } } > 4$ for all positive integers $n$.
\item Show that $u _ { n + 1 } < u _ { n }$ for $n \geqslant 1$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q3 [8]}}