A2 June 2019 Paper 1 Q4
4 Using the formulae for \(\displaystyle\sum_{r=1}^{n} r\) and \(\displaystyle\sum_{r=1}^{n} r^2\), show that \(\displaystyle\sum_{r=1}^{10} r(3r - 2) = 1045\). [3]
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\sum_{r=1}^{10} r(3r - 2) = \sum_{r=1}^{10} (3r^2 - 2r) = 3\sum_{r=1}^{10} r^2 - 2\sum_{r=1}^{10} r\) | M1 | 1.1 |
| \(= 3\left(\dfrac{1}{6}10.11.21\right) - 2\left(\dfrac{1}{2}10.11\right)\) | M1 | 2.1 |
| \((= 55(21 - 2) = 55 \times 19)\) \(= 1045\) | A1 | 1.1 |
| [3] |
Notes
M1: Separate soi
M1: Use both formulae with \(n = 10\)
A1: AG. oe 1155 − 110
Alternative method (leaving substitution to the end)
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{10} r(3r - 2) = \sum_{r=1}^{10} (3r^2 - 2r) = 3\sum_{r=1}^{10} r^2 - 2\sum_{r=1}^{10} r\) | M1 |
| \(= \dfrac{3}{6}n(n + 1)(2n + 1) - \dfrac{2}{2}n(n + 1)\) \(= \dfrac{1}{2}n(n + 1)(2n + 1 - 2) = \dfrac{1}{2}n(n + 1)(2n - 1)\) \(n = 10 \Rightarrow \dfrac{1}{2}10.11.19\) | M1 |
| \(= 1045\) | A1 |
| [3] |