AS June 2022 Paper 1 Q6
6
(a) Using standard summation formulae, show that \(\displaystyle\sum_{r=1}^{n} r(r + 2) = \tfrac{1}{6}n(n + 1)(2n + 7)\). [4]
(b) Use induction to prove the result in part (a). [6]
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\sum_{r=1}^{n} r(r + 2) = \sum_{r=1}^{n} r^2 + 2\sum_{r=1}^{n} r\) | M1 | 2.1 |
| \(= \dfrac{1}{6}n(n + 1)(2n + 1) + n(n + 1)\) | A1 | 1.1 |
| \(= \dfrac{1}{6}n(n + 1)(2n + 1 + 6)\) | M1 | 1.1 |
| \(= \dfrac{1}{6}n(n + 1)(2n + 7)\) | A1 | 2.3 |
| [4] |
Notes
M1: (2nd) Factoring \(n\) or \(n + 1\)
A1: (2nd) NB AG must show previous step
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\sum_{r=1}^{1} r(r + 2) = 1 \times 3 = 3,\ \dfrac{1}{6}n(n + 1)(2n + 7) = \dfrac{1}{6} \times 1 \times 2 \times 9 = 3\) so true for \(n = 1\) | B1 | 1.1 |
| Assume true for \(n = k\) \(\displaystyle\sum_{r=1}^{k+1} r(r + 2) = \dfrac{1}{6}k(k + 1)(2k + 7) + (k + 1)(k + 3)\) | M1 | 2.1 |
| \(= \dfrac{1}{6}(k + 1)(2k^2 + 7k + 6k + 18) = \dfrac{1}{6}(k + 1)(2k^2 + 13k + 18)\) | M1 | 1.1 |
| \(= \dfrac{1}{6}(k + 1)(k + 2)(2k + 9)\) | A1* | 2.1 |
| \(= \dfrac{1}{6}(k + 1)(k + 1 + 1)(2(k + 1) + 7)\) | A1 | 2.2a |
| so if true for \(n = k\) then true for \(n = k + 1\) as true for \(n = 1\), true for all \(n\) | A1cao | 2.4 |
| [6] |
Notes
B1: checking \(n = 1\)
M1: (1st) condone \(\displaystyle\sum_{r=1}^{k+1} k(k + 2) = \ldots\)
M1: (2nd) factorising
A1: (2nd) or equals target expression if given
A1cao: dep A1* must have both statements