FP1 June 2015 Q3
3.
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{n} (r + 1)(r + 4)\) | |
| \(= \displaystyle\sum_{r=1}^{n} r^2 + 5r + 4\) | B1 |
| \(= \tfrac{n}{6}(n + 1)(2n + 1) + 5\tfrac{n}{2}(n + 1) + 4n\) | M1 A1 |
| \(= \tfrac{n}{6}\{(n + 1)(2n + 1) + 15(n + 1) + 24\}\) | dM1 |
| \(= \tfrac{n}{6}\{(2n^2 + 3n + 1) + 15n + 15 + 24\}\) | |
| \(= \tfrac{n}{6}(2n^2 + 18n + 40)\) or \(= \tfrac{n}{3}(n^2 + 9n + 20)\) | |
| \(= \dfrac{n}{3}(n + 4)(n + 5)\) ** given answer** | A1* |
| (5) |
Notes
B1: Expands bracket correctly to \(r^2 + 5r + 4\)
M1: Uses \(\tfrac{n}{6}(n + 1)(2n + 1)\) or \(\tfrac{n}{2}(n + 1)\) correctly.
A1: Completely correct expression.
dM1: Attempts to remove factor \(\dfrac{n}{6}\) or \(\dfrac{n}{3}\) to obtain a quadratic factor. Need not be 3 term.
A1: Completely correct work including a step with a collected 3 term quadratic prior in the bracket with correct printed answer.
Accept approach which starts with LHS and then RHS which meet at \(\dfrac{n^3}{3} + 3n^2 + \dfrac{20n}{3}\). Award marks as above.
NB If induction attempted then typically this may only score the first B1.
However, consider the solution carefully and award as above if seen in the body of the induction attempt.
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=n+1}^{2n} (r + 1)(r + 4) = \frac{2n}{3}(2n + 4)(2n + 5) - \frac{n}{3}(n + 4)(n + 5)\) | M1 |
| \(= \dfrac{n}{3}\{8n^2 + 36n + 40 - n^2 - 9n - 20\}\) | dM1 |
| \(= \dfrac{n}{3}\{7n^2 + 27n + 20\} = \dfrac{n}{3}(n + 1)(7n + 20)\) or \(a = 7\), \(b = 20\) | A1 |
| (3) | |
| (8 marks) |
Notes
M1: Uses \(\mathrm{f}(2n) - \mathrm{f}(n)\) or \(\mathrm{f}(2n) - \mathrm{f}(n + 1)\) correctly. Require all 3 terms in \(2n\) (and \(n + 1\) if used).
dM1: Attempts to remove factor \(\dfrac{n}{6}\) or \(\dfrac{n}{3}\) to obtain a quadratic factor. Need not be 3 term.
A1: Either in expression or as above.