A2 October 2020 Paper 1 Q7
7 Prove by mathematical induction that \(\displaystyle\sum_{r=1}^{n} (r \times r!) = (n + 1)! - 1\) for all positive integers \(n\). [6]
| Scheme | Marks | AO |
|---|---|---|
| When \(n = 1\), \(\displaystyle\sum_{r=1}^{1} r \times r! = 1 \times 1! = 1 = 2! - 1\) | B1 | 1.1 |
| Assume true for \(n = k\) so \(\displaystyle\sum_{r=1}^{k} r \times r! = (k + 1)! - 1\) | M1 | 1.1 |
| then \(\displaystyle\sum_{r=1}^{k+1} r \times r! = (k + 1)! - 1 + (k + 1) \times (k + 1)!\) | M1 | 2.1 |
| \(= (k + 1)!(1 + k + 1) - 1 = (k + 2)! - 1 = (k + 1 + 1)! - 1\) | A1* | 2.1 |
| So if true for \(n = k\) then true for \(n = k + 1\) As true for \(n = 1\), true for all positive \(n\) | B1dep* B1 | 2.2a 2.4 |
| [6] |
Notes
A1*: Or target seen
B1: Final B mark only awarded if all previous marks awarded