A2 June 2019 Paper 1 Q1
1 Find \(\displaystyle\sum_{r=1}^{n} (2r^2 - 1)\), expressing your answer in fully factorised form. [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\sum_{r=1}^{n} (2r^2 - 1) = \frac{1}{3}n(n + 1)(2n + 1) - n\) | B1 B1 | 2.5 1.1b |
| \(= \dfrac{1}{3}n(2n^2 + 3n - 2)\) | B1 | 1.1b |
| \(= \dfrac{1}{3}n(2n - 1)(n + 2)\) | B1cao | 1.1b |
| [4] |
Notes
B1: \(\frac{1}{3}n(n + 1)(2n + 1)\ldots\)
B1: \(\ldots - n\)
B1: factoring out \(n\) correctly
B1cao: allow \(\frac{2}{6}n(2n - 1)(n + 2)\), etc