AS October 2021 Paper 1 Q7
7 Prove that \(2^{3n} - 3^n\) is divisible by 5 for all integers \(n \geqslant 1\). [5]
| Scheme | Marks | AO |
|---|---|---|
| Basis Case: when \(n = 1\): \(2^{3n} - 3^n = 2^3 - 3 = 8 - 3 = 5\) which is divisible by 5. | B1 | 2.1 |
| Assume true for \(n = k\) ie \(2^{3k} - 3^k = 5p\) for some integer \(p\) | M1 | 2.1 |
| \(2^{3(k+1)} - 3^{k+1} = 2^3 \times 2^{3k} - 3 \times 3^k\) \(= 8 \times (5p + 3^k) - 3 \times 3^k\) | M1 | 1.1 |
| \(= 5 \times 8p + 5 \times 3^k\) \(= 5(8p + 3^k) = 5q\) for some integer \(q\) and so this is also a multiple of 5 | A1 | 2.2a |
| So true for \(n = k \Rightarrow\) true for \(n = k + 1\). But true for \(n = 1\). So true for all integers \(n \geqslant 1\) | A1 | 2.4 |
| [5] |
Notes
B1: At least one intermediate step must be shown
M1: (1st) Must have statement in terms of some other variable than \(n\)
M1: (2nd) Uses laws of indices and then inductive hypothesis properly to eliminate either \(2^{3k}\) or \(3^k\) (not both)
or \(8 \times 2^{3k} - 3 \times (2^{3k} - 5p)\)
A1: (1st) AG. Further simplification to establish truth for \(k + 1\)
\(5(3p + 2^{3k})\)
A1: (2nd) Clear conclusion for induction process, following a correct proof by induction.
A formal proof by induction is required for full marks.