FP2 June 2006 Q2
2. Given that for all real values of \(r\), \[(2r + 1)^3 - (2r - 1)^3 = Ar^2 + B,\] where \(A\) and \(B\) are constants,
(a) find the value of \(A\) and the value of \(B\). (2)
(b) Hence, or otherwise, prove that \(\displaystyle\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n + 1)(2n + 1)\). (5)
(c) Calculate \(\displaystyle\sum_{r=1}^{40} (3r - 1)^2\). (3)
| Scheme | Marks |
|---|---|
| \((2r + 1)^3 = 8r^3 + 12r^2 + 6r + 1\) | |
| \((2r - 1)^3 = 8r^3 - 12r^2 + 6r - 1\) | |
| \((2r + 1)^3 - (2r - 1)^3 = 24r^2 + 2 \qquad (A = 24, B = 2)\) Accept \(r = 0 \Rightarrow B = 2\) and \(r = 1 \Rightarrow A + B = 26 \Rightarrow A = 24\) M1 for both | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\cancel{3^3} - 1^3 = 24 \times 1^2 + 2\) \(\cancel{5^3} - \cancel{3^3} = 24 \times 2^2 + 2\) \(\vdots\) \((2n + 1)^3 - \cancel{(2n - 1)^3} = 24 \times n^2 + 2\) | |
| \((2n + 1)^3 - 1^3 = 24\displaystyle\sum_{r=1}^{n} r^2 + \underline{2n}\) ft their B | M1 A1 A1ft |
| \(\displaystyle\sum_{r=1}^{n} r^2 = \frac{8n^3 + 12n^2 + 4n}{24}\) | M1 |
| \(= \dfrac{1}{6}n(2n^2 + 3n + 1) = \dfrac{1}{6}n(n + 1)(2n + 1)\) cso | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{40} (3r - 1)^2 = \sum_{r=1}^{40} (9r^2 - 6r + 1)\) | M1 |
| \(= 9 \times \dfrac{1}{6} \times 40 \times 41 \times 81 - 6 \times \dfrac{1}{2} \times 40 \times 41 + 40\) | M1 |
| \(= 194380\) | A1 |
| (3) | |
| (10 marks) |