AS June 2024 Paper 1 Q3
3
(a) Using standard summation formulae, write down an expression in terms of \(n\) for \(\displaystyle\sum_{r=1}^{2n} r^3\). [1]
(b) Hence show that \(\displaystyle\sum_{r=n+1}^{2n} r^3 = \tfrac{1}{4}n^2(an + b)(cn + d)\), where \(a\), \(b\), \(c\) and \(d\) are integers to be determined. [5]
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\sum_{r=1}^{2n} r^3 = \frac{1}{4}(2n)^2(2n + 1)^2\quad \left[= n^2(2n + 1)^2\right]\) | B1 | 1.1 |
| [1] |
Notes
B1: oe, \(\dfrac{1}{4}2n^2(2n + 1)^2\) is B0, mark final answer
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\sum_{r=n+1}^{2n} r^3 = \sum_{r=1}^{2n} r^3 - \sum_{r=1}^{n} r^3\) | M1 | 2.5 |
| \(= n^2(2n + 1)^2 - \dfrac{1}{4}n^2(n + 1)^2\) | A1 | 1.1 |
| \(= \dfrac{1}{4}n^2\left[4(2n + 1)^2 - (n + 1)^2\right]\) | M1 | 2.1 |
| \(= \dfrac{1}{4}n^2(4n + 2 + n + 1)(4n + 2 - n - 1)\) | A1 | 2.1 |
| \(= \dfrac{1}{4}n^2(5n + 3)(3n + 1)\) | A1 | 2.2a |
| [5] |
Notes
M1: sum from \(n + 1\) to \(2n\) is sum from 1 to \(2n\) − sum from 1 to \(n\)
M1: taking out common factor of \(n^2\) (at any stage)
A1: or \(\dfrac{1}{4}n^2(15n^2 + 14n + 3)\) oe