AS June 2024 Paper 1 Q12
12 Prove by induction that, for all \(n \in \mathbb{N}\), the expression
\[5^n - 2^n\]is divisible by 3 [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Writes \(5^1 - 2^1 = 3\) Accept \(5^0 - 2^0 = 0\) | B1 | 2.1 |
| Assumes \(5^k - 2^k\) is divisible by 3 and considers \(5^{k+1} - 2^{k+1}\) | M1 | 2.4 |
| Completes rigorous working to show that \(5^{k+1} - 2^{k+1}\) is divisible by 3 | A1 | 2.2a |
| Concludes a reasoned argument by stating that \(5^n - 2^n\) is divisible by 3 when \(n = 1\) (or \(n = 0\)), and that if \(5^n - 2^n\) is divisible by 3 when \(n = k\) then it is also divisible by 3 when \(n = k + 1\) and hence (by induction) \(5^n - 2^n\) is divisible by 3 for all \(n\) \((\in \mathbf{N})\) Condone reference to ‘rule’ / ‘statement’ in their final statement. Must define their \(m\) to be an integer. | R1 | 2.1 |
| (4 marks) |
Typical solution
\[5^1 - 2^1 = 3\]3 is divisible by 3
So the rule is true for \(n = 1\)
If the rule is true for \(n = k\) then \(5^k - 2^k = 3m\) for some integer \(m\)
\[\Rightarrow 5^k = 3m + 2^k\]\[\begin{aligned} 5^{k+1} - 2^{k+1} &= 5 \times 5^k - 2 \times 2^k \\ &= 5(3m + 2^k) - 2 \times 2^k \\ &= 15m + 5 \times 2^k - 2 \times 2^k \\ &= 15m + 3 \times 2^k \\ &= 3(5m + 2^k) \\ &= 3 \times \text{integer}\end{aligned}\]So the rule is also true for \(n = k + 1\)
Therefore, by induction
\(5^n - 2^n\) is divisible by 3 for all \(n \in \mathbf{N}\)