| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| 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. Part (a) is immediate substitution, (b) is a simple 'show that' requiring one more substitution, and (c) involves adding four terms and factorizing to show divisibility—all routine C1 techniques with no problem-solving insight needed. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \((a_2 =) 5k + 3\) | B1 | |
| (b) \((a_3 =) 5(5k + 3) + 3 = 25k + 18\) | M1 A1 cso | (*) |
| Notes | (a) \(5k + 3\) must be seen in (a) to gain the mark. (b) 1st M: Substitutes their \(a_2\) into \(5a_2 + 3\) - note the answer is given so working must be seen. (c) 1st M1: Substitutes their \(a_3\) into \(5a_3 + 3\) or uses \(125k + 93\). 2nd M1: for their sum \(k + a_2 + a_3 + a_4\) - must see evidence of four terms with plus signs and must not be sum of AP. 1st A1: All correct so far. 2nd A1ft: Limited ft – previous answer must be divisible by 6 (e.g \(156k + 42\)). This is dependent on second M mark in (c). Allow \(\frac{156k + 114}{6} = 26k + 19\) without explanation. No conclusion is needed. |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(a_4 = 5(25k + 18) + 3\) (= \(125k + 93\)) | M1 | |
| \(\sum a_r = k + (5k + 3) + (25k + 18) + (125k + 93)\) | M1 | |
| \(= 156k + 114\) | A1 | |
| \(= 6(26k + 19)\) (or explain each term is divisible by 6) | A1 ao | |
| (ii) | (4) / 7 |
(a) $(a_2 =) 5k + 3$ | B1 |
(b) $(a_3 =) 5(5k + 3) + 3 = 25k + 18$ | M1 A1 cso | (*)
Notes | | (a) $5k + 3$ must be seen in (a) to gain the mark. (b) 1st M: Substitutes their $a_2$ into $5a_2 + 3$ - note the answer is given so working must be seen. (c) 1st M1: Substitutes their $a_3$ into $5a_3 + 3$ or uses $125k + 93$. 2nd M1: for their sum $k + a_2 + a_3 + a_4$ - must see evidence of four terms with plus signs and must not be sum of AP. 1st A1: All correct so far. 2nd A1ft: Limited ft – previous answer must be divisible by 6 (e.g $156k + 42$). This is dependent on second M mark in (c). Allow $\frac{156k + 114}{6} = 26k + 19$ without explanation. No conclusion is needed.
(c)
(i) $a_4 = 5(25k + 18) + 3$ (= $125k + 93$) | M1 |
$\sum a_r = k + (5k + 3) + (25k + 18) + (125k + 93)$ | M1 |
$= 156k + 114$ | A1 |
$= 6(26k + 19)$ (or explain each term is divisible by 6) | A1 ao |
(ii) | | (4) / 7
---5. A sequence $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ is defined by
$$\begin{aligned}
a _ { 1 } & = k \\
a _ { n + 1 } & = 5 a _ { n } + 3 , \quad n \geqslant 1 ,
\end{aligned}$$
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 $a _ { 3 } = 25 k + 18$.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\sum _ { r = 1 } ^ { 4 } a _ { r }$ in terms of $k$, in its simplest form.
\item Show that $\sum _ { r = 1 } ^ { 4 } a _ { r }$ is divisible by 6 .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2011 Q5 [7]}}