A2 June 2025 Paper 1 Q3
3 Using standard summation formulae, show that, for integers \(n \geqslant 1\),
\(1 \times 3 + 2 \times 4 + \ldots + n \times (n + 2) = \dfrac{1}{6}n(n + 1)(an + b)\),
where \(a\) and \(b\) are integers to be determined. [5]
| Scheme | Marks | AO |
|---|---|---|
| \(1 \times 3 + 2 \times 4 + \ldots + n \times (n + 2) = \displaystyle\sum_{r=1}^{n} r(r + 2)\) | M1 | 2.1 |
| \(= \displaystyle\sum_{r=1}^{n} r^2 + 2\sum_{r=1}^{n} r\) | M1 | 2.5 |
| \(= \frac{1}{6}n(n + 1)(2n + 1) + 2 \times \frac{1}{2}n(n + 1)\) | M1 | 1.1 |
| \(= \frac{1}{6}n(n + 1)(2n + 1 + 6)\) | M1 | 1.1 |
| \(= \frac{1}{6}n(n + 1)(2n + 7)\) or \(a = 2, b = 7\) | A1 | 2.1 |
| [5] |
Notes
M1: converting series correctly to sigma notation
M1: writing their sum in terms of \(\sum_{r=1}^{n} r^2\) and \(\sum_{r=1}^{n} r\) soi
M1: substituting standard formulae for \(\sum_{r=1}^{n} r^2\) and \(\sum_{r=1}^{n} r\). \(r^2 + 2r\) or \(r(r + 2)\) must have been seen.
M1: correctly taking out a factor \(n\), \(n + 1\) or both from their expression; soi by correct final answer
A1: www but condone errors with sigma notation