| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Sequence defined by formula |
| Difficulty | Moderate -0.8 This is a straightforward application of the relationship between sum formulas and individual terms. Part (a) requires simple substitution (n=5), and part (b) uses the standard technique that a_n = S_n - S_{n-1}. Both are routine procedures with no conceptual difficulty or problem-solving required. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series |
| Answer | Marks | Guidance |
|---|---|---|
| (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 |
| (2 marks) | ||
| (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 \(12 + 4 \times 6^2 = ..., 12 + 4(6^2) = ... 12 + 24^2 = ...\) or indeed 156. You can accept the appearance of \(12 + 4 \times 6^2 = ...\) in a sum of terms. |
| \(a_6 = \sum_{r=1}^{6} a_r\) – their answer to 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 |
| (3 marks) | ||
| (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(a_n = (12 + 4n^2) - (12 + 4(n-1)^2) = 4(n^2 - (n-1)^2) = 4(2n-1)\) | M1 | Writes down an expression for \(a_n\) |
| \(a_6 = 4(2 \times 6 - 1) = ...\) | dM1 | Subs \(n = 6\) into the expression for \(a_n = 4(2n-1) = ...\) |
| cao 44 | A1 |
(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
| | **(2 marks)**
(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 $12 + 4 \times 6^2 = ..., 12 + 4(6^2) = ... 12 + 24^2 = ...$ or indeed 156. You can accept the appearance of $12 + 4 \times 6^2 = ...$ in a sum of terms.
$a_6 = \sum_{r=1}^{6} a_r$ – their answer to 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
| | **(3 marks)**
| | **(5 marks)**
**Alternative to 5(b)**
$a_n = (12 + 4n^2) - (12 + 4(n-1)^2) = 4(n^2 - (n-1)^2) = 4(2n-1)$ | M1 | Writes down an expression for $a_n$
$a_6 = 4(2 \times 6 - 1) = ...$ | dM1 | Subs $n = 6$ into the expression for $a_n = 4(2n-1) = ...$
cao 44 | A1 |
---Given that for all positive integers $n$,
$$\sum_{r=1}^{n} a_r = 12 + 4n^2$$
\begin{enumerate}[label=(\alph*)]
\item find the value of $\sum_{r=1}^{5} a_r$ [2]
\item Find the value of $a_6$ [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q5 [5]}}