FP1 June 2014 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 \[\sum_{r=1}^{n} (2r - 1)^2 = \frac{1}{3}n(4n^2 - 1)\] (6)
(b) Hence show that \[\sum_{r=2n+1}^{4n} (2r - 1)^2 = an(bn^2 - 1)\] where \(a\) and \(b\) are constants to be found. (3)
| Scheme | Marks |
|---|---|
| \(\big((2r - 1)^2 =\big)\, 4r^2 - 4r + 1\) | B1 |
| Proof by induction will usually score no more marks without use of standard results | |
| \(\displaystyle\sum_{r=1}^{n} (2r - 1)^2 = \sum_{r=1}^{n} (4r^2 - 4r + 1)\) | |
| \(= 4\sum r^2 - 4\sum r + \sum 1\) | |
| \(= \underline{4.\dfrac{1}{6}n(n + 1)(2n + 1) - 4.\dfrac{1}{2}n(n + 1)}, + n\) M1: An attempt to use at least one of the standard results correctly in summing at least 2 terms of their expansion of \((2r - 1)^2\) A1: Correct underlined expression oe B1: \(\sum 1 = n\) | M1A1B1 |
| \(= \dfrac{1}{3}n\left[4n^2 + 6n + 2 - 6n - 6 + 3\right]\) Attempt to factor out \(\dfrac{1}{3}n\) before given answer | M1 |
| \(= \dfrac{1}{3}n\left[4n^2 - 1\right]\) cso | A1 |
| (6) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=2n+1}^{4n} (2r - 1)^2 = \mathrm{f}(4n) - \mathrm{f}(2n) \text{ or } \mathrm{f}(2n + 1)\) Require some use of the result in part (a) for method. | M1 |
| \(= \dfrac{1}{3}4n\left(4.(4n)^2 - 1\right) - \dfrac{1}{3}.2n\left(4.(2n)^2 - 1\right)\) Correct expression | A1 |
| \(= \dfrac{2}{3}n\left[128n^2 - 2 - 16n^2 + 1\right]\) | |
| \(= \dfrac{2}{3}n\left[112n^2 - 1\right]\) Accept \(a = \dfrac{2}{3}, b = 112\) | A1 |
| (3) | |
| (9 marks) |