| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Recurrence relation: evaluate sum |
| Difficulty | Moderate -0.8 This is a straightforward recurrence relation question requiring only direct substitution to find terms and simple addition to evaluate the sum. Students need to work backwards once to find u₁ from u₂, then apply the formula twice forward, and finally add four numbers—all routine mechanical operations with no problem-solving or conceptual insight required. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_3 = 2u_2 - 1 = 2(9)-1 = 17\) | M1 | Substitutes 9 into RHS of iteration formula |
| \(u_4 = 2u_3 - 1 = 2(17)-1 = 33\) | A1 | Needs both \(u_3=17\) and \(u_4=33\) (but allow if either or both seen in part (b)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_1 = 5\) | B1 | For \(u_1=5\) however obtained; may appear in (a); may be called \(a=5\) |
| \(\sum_{r=1}^{4} u_r = 5 + 9 + 17 + 33 = 64\) | M1 | Uses their \(u_1\) found from \(u_2=2u_1-1\), adds to \(u_2\), their \(u_3\) and their \(u_4\) only |
| \(= 64\) | A1 | 64 |
## Question 4:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_3 = 2u_2 - 1 = 2(9)-1 = 17$ | M1 | Substitutes 9 into RHS of iteration formula |
| $u_4 = 2u_3 - 1 = 2(17)-1 = 33$ | A1 | Needs both $u_3=17$ and $u_4=33$ (but allow if either or both seen in part (b)) |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_1 = 5$ | B1 | For $u_1=5$ however obtained; may appear in (a); may be called $a=5$ |
| $\sum_{r=1}^{4} u_r = 5 + 9 + 17 + 33 = 64$ | M1 | Uses their $u_1$ found from $u_2=2u_1-1$, adds to $u_2$, their $u_3$ and their $u_4$ only |
| $= 64$ | A1 | 64 |
---4. A sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ satisfies
$$u _ { n + 1 } = 2 u _ { n } - 1 , n \geqslant 1$$
Given that $u _ { 2 } = 9$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $u _ { 3 }$ and the value of $u _ { 4 }$,
\item evaluate $\sum _ { r = 1 } ^ { 4 } u _ { r }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2013 Q4 [5]}}