| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Sigma notation: direct numerical evaluation |
| Difficulty | Moderate -0.8 This is a straightforward sigma notation question requiring only substitution and subtraction. Part (a) is direct substitution of n=5, while part (b) uses the standard technique that a₆ = S₆ - S₅. Both parts are routine applications of formulas with minimal problem-solving, making this easier than average for A-level. |
| Spec | 1.04g Sigma notation: for sums of series |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(\sum_{r=1}^{5} a_r = 12 + 4 \times 5^2 = ...\) | M1 |
| \(= 112\) | A1 | cao 112. Accept this answer with no incorrect working for both marks. If it is consequently summed it will be scored A0. |
| (b) | \(\sum_{r=1}^{6} a_r = 12 + 4 \times 6^2\) | M1 |
| \(a_6 = \sum_{r=1}^{6} a_r - \text{(part a)}\) | dM1 | Attempts to find their answer to \(\sum a_r\) – their answer to part \(a\). This is dependent upon the previous M mark. Also accept a restart where they attempt \(\sum_{r=1}^{6} a_r - \sum_{r=1}^{5} a_r\). |
| \(a_6 = 156 - 112 = 44\) | A1 | cao 44. |
| Alternative to 5(b): M1 Writes down an expression for \(a_n = (12 + 4n^2) - (12 + 4(n-1)^2) = 4(n^2 - (n-1)^2) = 4(2n - 1)\). dM1 Subs \(n = 6\) into the expression for \(a_n = 4(2n-1) = ...\). A1 cao 44. |
(a) | $\sum_{r=1}^{5} a_r = 12 + 4 \times 5^2 = ...$ | M1 | Substitutes $n=5$ into the expression $12 + 4n^2$ and attempt to find a numerical answer for $\sum a_r$. Accept as evidence expressions such as $12 + 4 \times 5^2 = ...$, $12 + 4(5)^2 = ...$, even $12 + 20^2 = 412$. Accept for this mark solutions which add $12 + 4 \times 1^2, 12 + 4 \times 2^2, 12 + 4 \times 3^2, 12 + 4 \times 4^2, 12 + 4 \times 5^2$ and as a result 112 appears in a sum.
| $= 112$ | A1 | cao 112. Accept this answer with no incorrect working for both marks. If it is consequently summed it will be scored A0.
(b) | $\sum_{r=1}^{6} a_r = 12 + 4 \times 6^2$ | M1 | Substitutes $n = 6$ into the expression $12 + 4n^2$. Accept as evidence $12 + 4 \times 6^2 = ..., 12 + 4(6^2) = ...$ or indeed 156. You can accept the appearance of $12 + 4 \times 6^2 = ...$ in a sum of terms.
| $a_6 = \sum_{r=1}^{6} a_r - \text{(part a)}$ | dM1 | Attempts to find their answer to $\sum a_r$ – their answer to part $a$. This is dependent upon the previous M mark. Also accept a restart where they attempt $\sum_{r=1}^{6} a_r - \sum_{r=1}^{5} a_r$.
| $a_6 = 156 - 112 = 44$ | A1 | cao 44.
| | | Alternative to 5(b): M1 Writes down an expression for $a_n = (12 + 4n^2) - (12 + 4(n-1)^2) = 4(n^2 - (n-1)^2) = 4(2n - 1)$. dM1 Subs $n = 6$ into the expression for $a_n = 4(2n-1) = ...$. A1 cao 44.
---5. Given that for all positive integers $n$,
$$\sum _ { r = 1 } ^ { n } a _ { r } = 12 + 4 n ^ { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item find the value of $\sum _ { r = 1 } ^ { 5 } a _ { r }$
\item Find the value of $a _ { 6 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2014 Q5 [5]}}