| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Recurrence relation: find parameter from given term |
| Difficulty | Moderate -0.8 This is a straightforward recurrence relation question requiring simple substitution to find u₂ and u₃, followed by solving a linear equation. It tests basic algebraic manipulation and understanding of sequence notation, but involves no problem-solving insight or challenging concepts—easier than the average A-level question. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series |
| Answer | Marks | Guidance |
|---|---|---|
| \(u_2 = 4k - 3\) | B1 | Allow \(4 \times k - 3\) or \(k \times 4 - 3\). |
| \(u_3 = 4(4k-3) - 3 = 16k - 15\) | M1 A1 (3 marks) | M1: uses iteration formula with their \(u_2 = 4k-3\) to find \(u_3 = 4 \times\)"\((4k-3)\)"\(-3\). A1: \(u_3 = 16k - 15\), must be simplified. Accept \(16 \times k - 15\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{n=1}^{3} u_n = k + 4k - 3 + 16k - 15\) | M1 | For knowing the sum equals \(k +\) "\(4k-3\)" \(+\) "\(16k-15\)". Attempt to write sum of their two terms and \(k\); addition/simplification need not be performed. |
| \(k + 4k - 3 + 16k - 15 = 18 \Rightarrow k = \ldots\) | dM1 | Sets their \(k + u_2 + u_3 = 18\) and attempts to solve for \(k\). Condone slips. |
| \(k = \frac{12}{7}\) | A1 (3 marks) | Or exact equivalent such as \(k = \frac{36}{21}\). |
## Question 4:
### Part (a):
$u_2 = 4k - 3$ | B1 | Allow $4 \times k - 3$ or $k \times 4 - 3$.
$u_3 = 4(4k-3) - 3 = 16k - 15$ | M1 A1 (3 marks) | M1: uses iteration formula with their $u_2 = 4k-3$ to find $u_3 = 4 \times$"$(4k-3)$"$-3$. A1: $u_3 = 16k - 15$, must be simplified. Accept $16 \times k - 15$.
### Part (b):
$\sum_{n=1}^{3} u_n = k + 4k - 3 + 16k - 15$ | M1 | For knowing the sum equals $k +$ "$4k-3$" $+$ "$16k-15$". Attempt to write sum of their two terms and $k$; addition/simplification need not be performed.
$k + 4k - 3 + 16k - 15 = 18 \Rightarrow k = \ldots$ | dM1 | Sets their $k + u_2 + u_3 = 18$ and attempts to solve for $k$. Condone slips.
$k = \frac{12}{7}$ | A1 (3 marks) | Or exact equivalent such as $k = \frac{36}{21}$.
SC in (b): candidates using $u_2 + u_3 + u_4 = 18 \Rightarrow k = \frac{99}{84}$ can be awarded SC 100.
---4. A sequence is defined by
$$\begin{aligned}
u _ { 1 } & = k , \text { where } k \text { is a constant } \\
u _ { n + 1 } & = 4 u _ { n } - 3 , n \geqslant 1
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item Find $u _ { 2 }$ and $u _ { 3 }$ in terms of $k$, simplifying your answers as appropriate.
Given $\sum _ { n = 1 } ^ { 3 } u _ { n } = 18$
\item find $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2019 Q4 [6]}}