FP1 June 2013 (R) Q10
10.
(i) Use the standard results for \(\displaystyle\sum_{r=1}^{n} r^3\) and \(\displaystyle\sum_{r=1}^{n} r\) to evaluate \[\sum_{r=1}^{24} (r^3 - 4r)\] (2)
(ii) Use the standard results for \(\displaystyle\sum_{r=1}^{n} r^2\) and \(\displaystyle\sum_{r=1}^{n} r\) to show that \[\sum_{r=0}^{n} (r^2 - 2r + 2n + 1) = \frac{1}{6}(n + 1)(n + a)(bn + c)\] for all integers \(n \geqslant 0\), where \(a\), \(b\) and \(c\) are constant integers to be found. (6)
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{24} (r^3 - 4r)\) | |
| \(= \dfrac{1}{4}24^2(24 + 1)^2 - 4.\dfrac{1}{2}24(24 + 1)\) An attempt to use at least one of the standard formulae correctly and substitute 24. | M1 |
| \(\{= 90000 - 1200\}\) \(= 88800\) 88800 | A1 cao |
| (2) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=0}^{n} (r^2 - 2r + 2n + 1)\) | |
| \(= \underline{\dfrac{1}{6}n(n + 1)(2n + 1) - 2.\dfrac{1}{2}n(n + 1)} + 2n(n + 1) + (n + 1)\) An attempt to use at least one of the standard formulae correctly. Correct underlined expression. \(2n \to 2n(n + 1)\) \(1 \to (n + 1)\) | M1 A1 B1 B1 |
| \(= \dfrac{1}{6}(n + 1)\{2n^2 + n - 6n + 12n + 6\}\) An attempt to factorise out \(\dfrac{1}{6}(n + 1)\) or \(\dfrac{1}{6}n\). | M1 |
| \(= \dfrac{1}{6}(n + 1)\{2n^2 + 7n + 6\}\) | |
| \(= \dfrac{1}{6}(n + 1)(n + 2)(2n + 3)\) Correct answer. (Note: \(a = 2, b = 2, c = 3\).) | A1 |
| (6) | |
| (8 marks) |