A2 June 2025 Paper 2 Q4
4 In this question you must show detailed reasoning.
Determine the sum of all cube numbers from 216 to 512 000 inclusive. [4]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\left(\sqrt[3]{512000} =\right) 80\) | B1 | 3.1a |
| \(\displaystyle\sum_{r=1}^{80} r^3 = \frac{1}{4} \times 80^2 \times (80 + 1)^2\) | M1 | 1.1 |
| \(= 10497600\) OR \(-\frac{1}{4} \times 5^2 \times 6^2\) OR \(-225\) | A1 | 1.1 |
| Required sum \(= 10497375\) | A1 | 2.2a |
| [4] |
Notes
B1: May be seen embedded in a sum
M1: Using the correct formula for sum of cubes with an upper limit (between 26-720 but not 216) substituted
A1: Any one of these seen correct
May see \(1 + 8 + 27 + 64 + 125 = 225\)
A1: Correct answer with no working is 0/4
Alternative method
| Scheme | Marks |
|---|---|
| DR \(\left(\sqrt[3]{512000} = 80\text{ and } 80 - 5 = \right) 75\) | B1 |
| Sum is \(\left(\displaystyle\sum_{r=6}^{80} r^3\right) = \sum_{R=1}^{75} (R + 5)^3 = \sum_{R=1}^{75} R^3 + 15R^2 + 75R + 125\) | M1 |
| \(= \frac{1}{4} \times 75^2 \times (75 + 1)^2\) \(+ 15 \times \frac{75(75 + 1)(2 \times 75 + 1)}{6} + 75 \times \frac{75(75 + 1)}{2}\) \(+ 125 \times 75\) | A1 |
| \(= 8122500 + 2151750 + 213750 + 9375\) \(= 10497375\) | A1 |
B1: May be seen embedded in a sum (e.g. 75 as upper limit)
M1: Expressing the required sum as a single sum and multiplying out so that standard formulae can be applied (allow one slip in coefficients). Ignore limits for this mark.
A1: 3 out of 4 terms correct (must include \(125 \times 75\))
A1: Correct answer with no working is 0/4