| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Recurrence relation: find specific terms |
| Difficulty | Standard +0.3 This is a straightforward C1 recurrence relation question requiring substitution to find expressions, then solving a quartic that factors nicely. While it involves multiple steps and algebraic manipulation, the techniques are routine (substitution, expanding brackets, solving equations) with no conceptual difficulty or novel insight required. Slightly above average due to the quartic equation, but the factorization is manageable. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.04e Sequences: nth term and recurrence relations |
| Answer | Marks |
|---|---|
| \(u_3 = (k^2-1)^2 - 1 = k^4 - 2k^2\) | B1 |
| M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(k^4 - k^2 - 12 = 0\) | M1 | |
| \((k^2+3)(k^2-4) = 0\) | M1 | |
| \(k^2 = -3\) (no solutions) or \(4\) | A1 | |
| \(k = \pm 2\) | A1 | (7 marks) |
(a) $u_2 = k^2 - 1$
$u_3 = (k^2-1)^2 - 1 = k^4 - 2k^2$ | B1 |
| M1 A1 |
(b) $k^4 - 2k^2 + k^2 - 1 = 11$
$k^4 - k^2 - 12 = 0$ | M1 |
$(k^2+3)(k^2-4) = 0$ | M1 |
$k^2 = -3$ (no solutions) or $4$ | A1 |
$k = \pm 2$ | A1 | (7 marks)The sequence $u_1, u_2, u_3, ...$ is defined by the recurrence relation
$$u_{n+1} = (u_n)^2 - 1, \quad n \geq 1.$$
Given that $u_1 = k$, where $k$ is a constant,
\begin{enumerate}[label=(\alph*)]
\item find expressions for $u_2$ and $u_3$ in terms of $k$. [3]
\end{enumerate}
Given also that $u_2 + u_3 = 11$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the possible values of $k$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q5 [7]}}