| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2017 |
| Session | Specimen |
| Marks | 6 |
| Topic | Sequences and series, recurrence and convergence |
| Type | Recurrence relation solving for closed form |
| Difficulty | Challenging +1.2 This is a Further Maths question on second-order linear recurrence relations with repeated roots (characteristic equation gives r=2 twice). Part (i) requires standard technique of finding the general solution u_n = (A + Bn)2^n and applying initial conditions. Part (ii) is straightforward once the closed form is found, as it's clear that (1-n)2^n is always an integer. While this is beyond standard A-level, it's a routine application of Further Maths techniques with no novel insight required. |
| Spec | 8.01g Second-order recurrence: solve with distinct, repeated, or complex roots |
\begin{enumerate}[label=(\roman*)]
\item Solve the recurrence relation $u_{n+2} = 4u_{n+1} - 4u_n$ for $n \geq 0$, given that $u_0 = 1$ and $u_1 = 1$. [4]
\item Show that each term of the sequence $\{u_n\}$ is an integer. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2017 Q4 [6]}}