A2 June 2023 Paper 1 Q8
8 Prove by mathematical induction that \(8^n - 3^n\) is divisible by 5 for all positive integers \(n\). [5]
| Scheme | Marks | AO |
|---|---|---|
| when \(n = 1\), \(8^n - 3^n = 5\) div by 5 | B1 | 2.1 |
| [Assume true when \(n = k\)] so \(8^k = 3^k + 5m\) | M1 | 2.1 |
| \(8^{k+1} - 3^{k+1} = 8(3^k + 5m) - 3 \times 3^k\) | M1 | 2.1 |
| \(= 5 \times 3^k + 40m = 5(3^k + 8m)\) div by 5 | A1 | 2.2a |
| so if true for \(n = k\) then true for \(n = k + 1\) As true for \(n = 1\), true for all \(n\). | A1 | 2.4 |
| [5] |
Notes
M1: (1st) or \(3^k = 8^k - 5m\) or \(8^k - 3^k = 5m\) or \(8^k - 3^k\) div 5
M1: (2nd) assumption used
A1: (1st) successful completion
\(3^k = 8^k - 5m\) used \(\Rightarrow 5(8^k + 3m)\) div 5 www
\(8^k - 3^k = 5m\) used \(\Rightarrow 5(8m + 3^k)\) div 5 www
A1: (2nd) must receive all previous marks for this to be awarded
Alternative method
| Scheme | Marks |
|---|---|
| when \(n = 1\), \(8^n - 3^n = 5\) div by 5 | B1 |
| [Assume true when \(n = k\)] so \(u_k = 8^k - 3^k = 5m\) | M1 |
| \(u_{k+1} - u_k = 8^{k+1} - 3^{k+1} - (8^k - 3^k)\) \(= 2(8^k - 3^k) + 5(8^k)\) | M1 |
| \(= 2(5m) + 5(8^k) = 5(2m + 8^k)\) div by 5 | A1 |
| so if true for \(n = k\) then true for \(n = k + 1\) As true for \(n = 1\), true for all \(n\). | A1 |
M1: (2nd) considering difference between \(u_{k+1}\) and \(u_k\)
A1: (2nd) must receive all previous marks for this to be awarded