| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2005 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Non-linear or complex iterative formula convergence |
| Difficulty | Challenging +1.2 This is a Further Maths question involving proof by induction on a recurrence relation and linear approximation of convergence behavior. Part (i) requires a standard induction proof with straightforward algebra showing u_n < 2 implies u_{n+1} < 2. Part (ii) involves substituting u_n = 2 - ε into the recurrence and using binomial approximation (1+x)^{1/2} ≈ 1 + x/2 for small x, which is a routine technique at this level. While it requires multiple techniques and careful algebraic manipulation, both parts follow standard Further Maths patterns without requiring novel insight. |
| Spec | 4.01a Mathematical induction: construct proofs8.01c Sequence behaviour: periodic, convergent, divergent, oscillating, monotonic |
The sequence $u_1, u_2, u_3, \ldots$ is such that $u_1 = 1$ and
$$u_{n+1} = -1 + \sqrt{(u_n + 7)}.$$
\begin{enumerate}[label=(\roman*)]
\item Prove by induction that $u_n < 2$ for all $n \geqslant 1$. [4]
\item Show that if $u_n = 2 - \varepsilon$, where $\varepsilon$ is small, then
$$u_{n+1} \approx 2 - \frac{1}{6}\varepsilon.$$ [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP1 2005 Q2 [6]}}