FP1 June 2013 Q5
5.
(a) Use the standard results for \(\displaystyle\sum_{r=1}^{n} r\) and \(\displaystyle\sum_{r=1}^{n} r^2\) to show that \[\sum_{r=1}^{n} (r + 2)(r + 3) = \frac{1}{3}n(n^2 + 9n + 26)\] for all positive integers \(n\). (6)
(b) Hence show that \[\sum_{r=n+1}^{3n} (r + 2)(r + 3) = \frac{2}{3}n(an^2 + bn + c)\] where \(a\), \(b\) and \(c\) are integers to be found. (4)
| Scheme | Marks |
|---|---|
| \((r + 2)(r + 3) = r^2 + 5r + 6\) | B1 |
| \(\sum\left(r^2 + 5r + 6\right) = \dfrac{1}{6}n(n + 1)(2n + 1) + 5 \times \dfrac{1}{2}n(n + 1),\ +6n\) M1: Use of correct expressions for \(\sum r^2\) and \(\sum r\) B1ft: \(\sum k = nk\) | M1,B1ft |
| \(= \dfrac{1}{3}n\left[\dfrac{1}{2}(n + 1)(2n + 1) + \dfrac{15}{2}(n + 1) + 18\right]\) M1: Factors out \(n\) ignoring treatment of constant. A1: Correct expression with \(\dfrac{1}{3}n\) or \(\dfrac{1}{6}n\) factored out, allow recovery. | M1 A1 |
| \(\left(= \dfrac{1}{3}n\left[n^2 + \dfrac{3}{2}n + \dfrac{1}{2} + \dfrac{15}{2}n + \dfrac{15}{2} + 18\right]\right)\) \(= \dfrac{1}{3}n\left[n^2 + 9n + 26\right]\) * Correct completion to printed answer | A1*cso |
| (6) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=n+1}^{3n} = \dfrac{1}{3}3n\left((3n)^2 + 9(3n) + 26\right) - \dfrac{1}{3}n\left(n^2 + 9n + 26\right)\) M1: \(\mathrm{f}(\mathbf{3n}) - \mathrm{f}(n \text{ or } n + 1)\) and attempt to use part (a). A1: Equivalent correct expression | M1A1 |
| \(\left(= n\left(9n^2 + 27n + 26\right) - \dfrac{1}{3}n\left(n^2 + 9n + 26\right)\right)\) | |
| \(= \dfrac{2}{3}n\left(\dfrac{27}{2}n^2 + \dfrac{81}{2}n + 39 - \dfrac{1}{2}n^2 - \dfrac{9}{2}n - 13\right)\) Factors out \(= \dfrac{2}{3}n\) dependent on previous M1 | dM1 |
| \(= \dfrac{2}{3}n\left(13n^2 + 36n + 26\right)\) Accept correct expression. | A1 |
| \((a = 13, b = 36, c = 26)\) | |
| (4) | |
| Total 10 |
Notes
\(\mathbf{3}\mathrm{f}(\boldsymbol{n}) - \mathrm{f}(n \text{ or } n + 1)\) is M0