FP1 June 2012 Q4
4.
(a) Use the standard results for \(\displaystyle\sum_{r=1}^{n} r^3\) and \(\displaystyle\sum_{r=1}^{n} r\) to show that \[\sum_{r=1}^{n} \left(r^3 + 6r - 3\right) = \frac{1}{4}n^2(n^2 + 2n + 13)\] for all positive integers \(n\). (5)
(b) Hence find the exact value of \[\sum_{r=16}^{30} \left(r^3 + 6r - 3\right)\] (2)
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{n} \left(r^3 + 6r - 3\right)\) | |
| \(= \underline{\dfrac{1}{4}n^2(n + 1)^2 + 6.\dfrac{1}{2}n(n + 1)} - 3n\) M1; An attempt to use at least one of the standard formulae correctly in summing at least 2 terms of \(r^3 + 6r - 3\) A1: Correct underlined expression. B1: \(-3 \to -3n\) | M1A1B1 |
| \(= \dfrac{1}{4}n^2(n + 1)^2 + 3n^2 + 3n - 3n\) | |
| \(= \dfrac{1}{4}n^2(n + 1)^2 + 3n^2\) \(= \dfrac{1}{4}n^2\left((n + 1)^2 + 12\right)\) Cancels out the \(3n\) and attempts to factorise out at least \(\dfrac{1}{4}n\). | dM1 |
| \(= \dfrac{1}{4}n^2\left(n^2 + 2n + 13\right)\) (AG) Correct answer with no errors seen. | A1 * |
| [5] |
Notes
If any marks have been lost, no further marks are available in part (a)
Provided the first 3 marks are scored, allow the next two marks for correctly showing the algebraic equivalence. E.g. showing that both \(\dfrac{1}{4}n^2(n + 1)^2 + 6.\dfrac{1}{2}n(n + 1) - 3n\) and \(\dfrac{1}{4}n^2\left(n^2 + 2n + 13\right) = \dfrac{1}{4}n^4 + \dfrac{1}{2}n^3 + \dfrac{13}{4}n^2\)
There are no marks for proof by induction but apply the scheme if necessary.
Alternative (Way 2)
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{n} \left(r^3 + 6r - 3\right)\) | |
| \(= \underline{\dfrac{1}{4}n^2(n + 1)^2 + 6.\dfrac{1}{2}n(n + 1)} - 3n\) An attempt to use at least one of the standard formulae correctly. Correct underlined expression. \(-3 \to -3n\) | M1 A1 B1 |
| If any marks have been lost, no further marks are available in part (a). | |
| \(= \dfrac{1}{4}n\left(n(n + 1)^2 + 12(n + 1) - 12\right)\) \(= \dfrac{1}{4}n\left(n(n + 1)^2 + 12n + 12 - 12\right)\) \(= \dfrac{1}{4}n\left(n(n + 1)^2 + 12n\right)\) Attempts to factorise out at least \(\dfrac{1}{4}n\) from a correct expression and cancels the constant inside the brackets. | dM1 |
| \(= \dfrac{1}{4}n^2\left(n^2 + 2n + 13\right)\) (AG) Correct answer | A1 * |
| [5] |
| Scheme | Marks |
|---|---|
| \(S_n = \displaystyle\sum_{r=16}^{30} \left(r^3 + 6r - 3\right) = S_{30} - S_{15}\) | |
| \(= \tfrac{1}{4}(30)^2\left(30^2 + 2(30) + 13\right) - \tfrac{1}{4}(15)^2\left(15^2 + 2(15) + 13\right)\) Use of \(S_{30} - S_{15}\) or \(S_{30} - S_{16}\) | M1 |
| \(= 218925 - 15075\) | |
| \(= 203850\) 203850 | A1 cao |
| [2] | |
| 7 marks |
Notes
NB They must be using \(S_n = \dfrac{1}{4}n^2\left(n^2 + 2n + 13\right)\) not \(S_n = n^3 + 6n - 3\)
NB \(S_{30} - S_{16} = 218925 - 19264 = 199661\) (Scores M1 A0)