A2 October 2020 Paper 1 Q7
7 Prove by induction that the sum of the cubes of three consecutive positive integers is divisible by 9. [5]
| Scheme | Marks | AO |
|---|---|---|
| Basis case: \(1^3 + 2^3 + 3^3 = 36 = 4 \times 9\) | B1 | 2.1 |
| Consider the sum \(\mathrm{f}(r) = r^3 + (r + 1)^3 + (r + 2)^3\) Assume that \(\mathrm{f}(r) = 9k\) for some \(k \in \mathbb{Z}\) Then \(\mathrm{f}(r + 1) = \mathrm{f}(r) + (r + 3)^3 - r^3\) | M1 | 2.1 |
| \(= \mathrm{f}(r) + r^3 + 3r^2 \times 3 + 3r \times 9 + 27 - r^3\) \(= \mathrm{f}(r) + 9r^2 + 27r + 27\) | M1 | 2.1 |
| \(= 9k + 9\left(r^2 + 3r + 3\right) = 9k'\) for some \(k' \in \mathbb{Z}\) | A1 | 2.2a |
| So if true for \(r\) then true also for \(r + 1\) But it is true for \(r = 1\) so is true for all integers \(r\) | A1 | 2.5 |
| [5] |
Notes
B1: Basis case – either \(4 \times 9\) or “36 is divisible by 9” given explicitly.
M1: For getting started (Sight of \(\mathrm{f}(r)\) is not necessary)
M1: For finding \(\mathrm{f}(r + 1)\)
A1: Each term as a multiple of 9
A1: Conclusion dependant on all other marks being earned