FP1 June 2017 Q8
8.
(a) Use the standard results for \(\displaystyle\sum_{r=1}^{n} r\) and \(\displaystyle\sum_{r=1}^{n} r^2\) to show that \[\sum_{r=1}^{n}(3r^2 + 8r + 3) = \frac{1}{2}n(2n + 5)(n + 3)\] for all positive integers \(n\). (5)
Given that \[\sum_{r=1}^{12}\left(3r^2 + 8r + 3 + k(2^{r-1})\right) = 3520\]
(b) find the exact value of the constant \(k\). (4)
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{n}\left(3r^2 + 8r + 3\right)\) | |
| \(= \dfrac{3}{6}n(n + 1)(2n + 1) + \dfrac{8}{2}n(n + 1) + 3n\) | M1 A1 B1 |
| \(= \dfrac{1}{2}n(n + 1)(2n + 1) + 4n(n + 1) + 3n\) | |
| \(= \dfrac{1}{2}n\left((2n + 1)(n + 1) + 8(n + 1) + 6\right)\) | M1 |
| \(= \dfrac{1}{2}n\left(2n^2 + 3n + 1 + 8n + 8 + 6\right)\) | |
| \(= \dfrac{1}{2}n\left(2n^2 + 11n + 15\right)\) | |
| \(= \dfrac{1}{2}n(2n + 5)(n + 3)\quad (*)\) | A1*cso |
| (5) |
Notes
M1: An attempt to use at least one of the correct standard formulae for first two terms.
A1: Correct first two terms.
B1: \(3 \to 3n\)
M1: Factorise out at least \(n\) from all terms at any point. There must be a factor of \(n\) in every term.
A1*cso: Achieves the correct answer, no errors seen.
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{12}\left(3r^2 + 8r + 3 + k(2^{r-1})\right) = 3520\) | |
| \(\displaystyle\sum_{r=1}^{12}\left(3r^2 + 8r + 3\right) = \frac{1}{2}(12)(29)(15)\ \{= 2610\}\) | M1 |
| \(\displaystyle\sum_{r=1}^{12}(2^{r-1}) = \frac{1(1 - 2^{12})}{1 - 2}\ \{= 4095\}\) | M1 A1 |
| So, \(2610 + 4095k = 3520 \Rightarrow 4095k = 910\) | |
| giving, \(k = \dfrac{2}{9}\) | A1 |
| (4) | |
| (9 marks) |
Notes
M1: Attempt to evaluate \(\displaystyle\sum_{r=1}^{12}\left(3r^2 + 8r + 3\right)\)
M1: Attempt to apply the sum to 12 terms of a GP or adds up all 12 terms.
A1: \(\dfrac{1(1 - 2^{12})}{1 - 2}\) o.e. or 4095.
A1: \(k = \dfrac{2}{9}\) or \(0.\dot{2}\)
Note: 2nd M1 1st A1: These two marks can be implied by seeing 4095 or \(4095k\)