AS June 2021 Paper 1 Q9
9
(a) Use the standard formulae for \(\displaystyle\sum_{r=1}^{n} r\) and \(\displaystyle\sum_{r=1}^{n} r^2\) to show that\[\sum_{r=1}^{n} r(r + 3) = an(n + 1)(n + b)\]
where \(a\) and \(b\) are constants to be determined. [4 marks]
(b) Hence, or otherwise, find a fully factorised expression for\[\sum_{r=n+1}^{5n} r(r + 3)\]
[3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Writes the sum in the form \(\alpha\sum r^2 + \beta\sum r\) where \(\alpha\) and \(\beta\) are numbers. PI by use of correct standard formulae. | M1 | 1.1a |
| Recalls and uses standard formulae to obtain a correct expression for the sum in terms of \(n\) May be unsimplified. | A1 | 1.2 |
| Identifies \(n\) and \((n + 1)\) as common factors. Must be at least one correct term inside the remaining bracket. | M1 | 1.1a |
| Completes a fully correct proof to reach the required result, including a clear statement that the original sum can be written in the form \(\alpha\sum r^2 + \beta\sum r\) Must have \(a = \frac{1}{3}\) and \(b = 5\) | R1 | 2.1 |
| (4) |
Typical solution
\[\begin{aligned}\sum_{r=1}^{n} r(r + 3) &= \sum_{r=1}^{n} r^2 + 3\sum_{r=1}^{n} r \\ &= \frac{1}{6}n(n + 1)(2n + 1) + 3 \times \frac{1}{2}n(n + 1) \\ &= \frac{1}{6}n(n + 1)\big((2n + 1) + 9\big) \\ &= \frac{1}{6}n(n + 1)(2n + 10) \\ &= \frac{1}{3}n(n + 1)(n + 5)\end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Substitutes \(5n\) into their expression of the form \(an(n + 1)(n + b)\) and subtracts their part (a). | M1 | 1.1a |
| Obtains a correct expression. May be unsimplified. | A1 | 1.1b |
| Obtains a correct fully factorised expression. | A1 | 1.1b |
| (3) | ||
| (7 marks) |