| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| 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 and basic algebraic manipulation. Parts (a) and (b) involve simple iteration of the formula, part (c)(i) requires adding four terms, and part (c)(ii) is a simple divisibility check by factoring. No problem-solving insight needed—purely mechanical application of the given recurrence relation. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series |
(a) B1 for $(a =) 3k + 5$ [must be seen in part (a) or labelled $a = $]
(b) M1 for $(a =) 3(3k+5) + 5$
$= 9k + 20$ (*)
A1cso
(c)(i) M1 for $a = 3(9k+20) + 5$ ($ = 27k + 65$)
M1 for $\sum_{r=1}^{4} a_r = k + (3k+5) + (9k+20) + (27k+65)$
A1 for $= 40k + 90$
A1ft for $= 10(4k+9)$ (or explain why divisible by 10)
---8. A sequence $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ is defined by
$$\begin{aligned}
a _ { 1 } & = k \\
a _ { n + 1 } & = 3 a _ { n } + 5 , \quad n \geqslant 1
\end{aligned}$$
where $k$ is a positive integer.
\begin{enumerate}[label=(\alph*)]
\item Write down an expression for $a _ { 2 }$ in terms of $k$.
\item Show that $a _ { 3 } = 9 k + 20$.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\sum _ { r = 1 } ^ { 4 } a _ { r }$ in terms of $k$.
\item Show that $\sum _ { r = 1 } ^ { 4 } a _ { r }$ is divisible by 10 .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2007 Q8 [7]}}