AS October 2020 Paper 1 Q5
5 You are given that \(u_1 = 5\) and \(u_{n+1} = u_n + 2n + 4\).
Prove by induction that \(u_n = n^2 + 3n + 1\) for all positive integers \(n\). [6]
| Scheme | Marks | AO |
|---|---|---|
| When \(n = 1\), \(n^2 + 3n + 1 = 1^2 + 3 \times 1 + 1 = 5\) | B1 | 1.1 |
| Assume true for \(n = k\) \(u_k = k^2 + 3k + 1\) | B1 | 2.1 |
| Then \(u_{k+1} = u_k + 2k + 4\) \(= k^2 + 3k + 1 + 2k + 4\) | M1 | 1.1 |
| \(= (k + 1)^2 + 3k + 4\) \(= (k + 1)^2 + 3(k + 1) + 1\) | A1* | 2.1 |
| So if true for \(n = k\), then true for \(n = k + 1\) | B1*dep | 2.2a |
| As true for \(n = 1\), true for all \(n\). | B1cao | 2.4 |
| [6] |
Notes
B1*dep: dep A1*