C1 January 2014 (IAL) Q5
5. Given that for all positive integers \(n\),\[\sum_{r=1}^{n} a_r = 12 + 4n^2\]
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{5} a_r = 12 + 4\times 5^2 = ..\) | M1 |
| \(= 112\) | A1 |
| (2) |
Notes
M1: Substitutes \(n\)=5 into the expression \(12 + 4n^2\) and attempt to find a numerical answer for \(\displaystyle\sum_{r=1}^{5} a_r\).
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.
A1: cao 112. Accept this answer with no incorrect working for both marks. If it is consequently summed it will be scored A0
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{6} a_r = 12 + 4\times 6^2\) | M1 |
| \(a_6 = \displaystyle\sum_{r=1}^{r=6} a_r - (\text{part } a)\) | dM1 |
| \(a_6 = 156 - 112 = 44\) | A1 |
| (3) | |
| (5 marks) |
Notes
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.
dM1: Attempts to find their answer to \(\displaystyle\sum_{r=1}^{6} a_r\) – their answer to part \(a\)
This is dependent upon the previous M mark.
Also accept a restart where they attempt \(\displaystyle\sum_{r=1}^{6} a_r - \sum_{r=1}^{5} a_r\)
A1: cao 44
Alternative to 5(b)
M1: Writes down an expression for \(a_n = \left(12 + 4n^2\right) - \left(12 + 4(n - 1)^2\right) = 4\left(n^2 - (n - 1)^2\right) = 4(2n - 1)\)
dM1: Subs \(n = 6\) into the expression for \(a_n = 4(2n - 1) = \ldots\)
A1: cao 44