| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Complete square then solve equation |
| Difficulty | Moderate -0.8 This is a straightforward substitution and solving of a quadratic equation. Students set n² - 6n + 11 = 38, rearrange to n² - 6n - 27 = 0, and factor or use the quadratic formula. It's simpler than average A-level questions as it requires only basic algebraic manipulation with no conceptual challenges or multi-step reasoning. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| \(u_k = k^2 - 6k + 11 = 38\) | M1 | |
| \(\therefore k^2 - 6k - 27 = 0\) | M1 | |
| \((k+3)(k-9) = 0\) | M1 | |
| \(k \geq 1 \therefore k = 9\) | A1 | (3) |
$u_k = k^2 - 6k + 11 = 38$ | M1 |
$\therefore k^2 - 6k - 27 = 0$ | M1 |
$(k+3)(k-9) = 0$ | M1 |
$k \geq 1 \therefore k = 9$ | A1 | (3)\begin{enumerate}
\item The $n$th term of a sequence is defined by
\end{enumerate}
$$u _ { n } = n ^ { 2 } - 6 n + 11 , \quad n \geq 1 .$$
Given that the $k$ th term of the sequence is 38 , find the value of $k$.\\
\hfill \mbox{\textit{Edexcel C1 Q1 [3]}}