FP1 June 2013 (R) Q8
8.
(a) Prove by induction, that for \(n \in \mathbb{Z}^{+}\), \[\sum_{r=1}^{n} r(2r - 1) = \frac{1}{6}n(n + 1)(4n - 1)\] (6)
(b) Hence, show that \[\sum_{r=n+1}^{3n} r(2r - 1) = \frac{1}{3}n(an^2 + bn + c)\] where \(a\), \(b\) and \(c\) are integers to be found. (4)
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{n} r(2r - 1) = \frac{1}{6}n(n + 1)(4n - 1)\) | |
| \(n = 1\); \(\quad \text{LHS} = \displaystyle\sum_{r=1}^{1} r(2r - 1) = 1\) \(\phantom{n = 1; \quad} \text{RHS} = \tfrac{1}{6}(1)(2)(3) = 1\) As LHS = RHS, the summation formula is true for \(n = 1\). \(\tfrac{1}{6}(1)(2)(3) = 1\) seen | B1 |
| Assume that the summation formula is true for \(n = k\). ie. \(\displaystyle\sum_{r=1}^{k} r(2r - 1) = \frac{1}{6}k(k + 1)(4k - 1)\). | |
| With \(n = k + 1\) terms the summation formula becomes: \(\displaystyle\sum_{r=1}^{k+1} r(2r - 1) = \underline{\frac{1}{6}k(k + 1)(4k - 1) + (k + 1)(2(k + 1) - 1)}\) \(S_{k+1} = S_k + u_{k+1}\) with \(S_k = \dfrac{1}{6}k(k + 1)(4k - 1)\). | M1 |
| \(= \dfrac{1}{6}k(k + 1)(4k - 1) + (k + 1)(2k + 1)\) | |
| \(= \dfrac{1}{6}(k + 1)\big(k(4k - 1) + 6(2k + 1)\big)\) Factorise by \(\dfrac{1}{6}(k + 1)\) | dM1 |
| \(= \dfrac{1}{6}(k + 1)(4k^2 + 11k + 6)\) \((4k^2 + 11k + 6)\) or equivalent quadratic seen | A1 |
| \(= \dfrac{1}{6}(k + 1)(k + 2)(4k + 3)\) | |
| \(= \dfrac{1}{6}(k + 1)(k + 1 + 1)(4(k + 1) - 1)\) Correct completion to \(S_{k+1}\) in terms of \(k + 1\) dependent on both Ms. | dM1 |
| If the summation formula is true for \(n = k\), then it is shown to be true for \(n = k + 1\). As the result is true for \(n = 1\), it is now also true for all \(n\) and \(n \in \mathbb{Z}^{+}\) by mathematical induction. Conclusion with all 4 underlined elements that can be seen anywhere in the solution | A1 cso |
| (6) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=n+1}^{3n} r(2r - 1) = \mathrm{S}_{3n} - \mathrm{S}_{n}\) | |
| \(= \dfrac{1}{6}.3n(3n + 1)(12n - 1) - \dfrac{1}{6}n(n + 1)(4n - 1)\) Use of \(\mathrm{S}_{3n} - \mathrm{S}_{n}\) or \(\mathrm{S}_{3n} - \mathrm{S}_{n+1}\) with the result from (a) used at least once. Correct un-simplified expression. | M1 A1 |
| \(= \dfrac{1}{6}n\{3(3n + 1)(12n - 1) - (n + 1)(4n - 1)\}\) | |
| \(= \dfrac{1}{6}n\{3(36n^2 + 9n - 1) - (4n^2 + 3n - 1)\}\) Factorises out \(\dfrac{1}{6}n\) or \(\dfrac{1}{3}n\) and an attempt to open up the brackets. | dM1 |
| \(= \dfrac{1}{6}n\{108n^2 + 27n - 3 - 4n^2 - 3n + 1\}\) | |
| \(= \dfrac{1}{6}n\{104n^2 + 24n - 2\}\) | |
| \(= \dfrac{1}{3}n(52n^2 + 12n - 1)\) \(\{a = 52, b = 12, c = -1\}\) \(= \dfrac{1}{3}n(52n^2 + 12n - 1)\) | A1 |
| (4) | |
| (10 marks) |