FP1 June 2009 Q2
2.
(a) Using the formulae for \(\displaystyle\sum_{r=1}^{n} r\), \(\displaystyle\sum_{r=1}^{n} r^2\) and \(\displaystyle\sum_{r=1}^{n} r^3\), show that \[\sum_{r=1}^{n} r(r + 1)(r + 3) = \frac{1}{12}n(n + 1)(n + 2)(3n + k),\] where \(k\) is a constant to be found. (7)
(b) Hence evaluate \(\displaystyle\sum_{r=21}^{40} r(r + 1)(r + 3)\). (2)
| Scheme | Marks |
|---|---|
| \(r(r + 1)(r + 3) = r^3 + 4r^2 + 3r\), so use \(\sum r^3 + 4\sum r^2 + 3\sum r\) | M1 |
| \(= \dfrac{1}{4}n^2(n + 1)^2 + 4\left(\dfrac{1}{6}n(n + 1)(2n + 1)\right) + 3\left(\dfrac{1}{2}n(n + 1)\right)\) | A1 A1 |
| \(= \dfrac{1}{12}n(n + 1)\{3n(n + 1) + 8(2n + 1) + 18\}\) or \(= \dfrac{1}{12}n\{3n^3 + 22n^2 + 45n + 26\}\) or \(= \dfrac{1}{12}(n + 1)\{3n^3 + 19n^2 + 26n\}\) | M1 A1 |
| \(= \dfrac{1}{12}n(n + 1)\{3n^2 + 19n + 26\} = \dfrac{1}{12}n(n + 1)(n + 2)(3n + 13)\) \((k = 13)\) | M1 A1cao |
| (7) |
Notes
(a) M1 expand and must start to use at least one standard formula
First 2 A marks: One wrong term A1 A0, two wrong terms A0 A0.
M1: Take out factor \(kn(n + 1)\) or \(kn\) or \(k(n + 1)\) directly or from quartic
A1: See scheme (cubics must be simplified)
M1: Complete method including a quadratic factor and attempt to factorise it
A1 Completely correct work.
Just gives \(k = 13\), no working is 0 marks for the question.
Alternative method
Expands \((n + 1)(n + 2)(3n + k)\) and confirms that it equals \(\{3n^3 + 22n^2 + 45n + 26\}\) together with statement \(k = 13\) can earn last M1A1
The previous M1A1 can be implied if they are using a quartic.
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{21}^{40} = \sum_{1}^{40} - \sum_{1}^{20}\) | M1 |
| \(= \dfrac{1}{12}(40 \times 41 \times 42 \times 133) - \dfrac{1}{12}(20 \times 21 \times 22 \times 73) = 763420 - 56210 = 707210\) | A1 cao |
| (2) | |
| [9] |
Notes
(b) M1 is for substituting 40 and 20 into their answer to (a) and subtracting. (NB not 40 and 21)Adding terms is M0A0 as the question said “Hence”