FP1 June 2016 Q3
3.
(a) Using the formula for \(\displaystyle\sum_{r=1}^{n} r^2\) write down, in terms of \(n\) only, an expression for \[\sum_{r=1}^{3n} r^2\] (1)
(b) Show that, for all integers \(n\), where \(n > 0\) \[\sum_{r=2n+1}^{3n} r^2 = \frac{n}{6}(an^2 + bn + c)\] where the values of the constants \(a\), \(b\) and \(c\) are to be found. (4)
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{3n} r^2 = \frac{1}{6}3n(3n + 1)(6n + 1)\) or \(\displaystyle\sum_{r=1}^{3n} r^2 = \frac{1}{2}n(3n + 1)(6n + 1)\) or equivalent | B1 |
| (1) |
Notes
B1: Either right hand side or exact equivalent - isw if expanded
| Scheme | Marks |
|---|---|
| See \(\displaystyle\sum_{r=1}^{2n} r^2 = \frac{1}{3}n(2n + 1)(4n + 1)\) or equivalent | B1 |
| Attempt to use \(\displaystyle\sum_{r=1}^{3n} r^2 - \sum_{r=1}^{2n} r^2 = \frac{n}{6}\{3(3n + 1)(6n + 1) - 2(2n + 1)(4n + 1)\}\) | M1 |
| \(= \dfrac{n}{6}\{(54n^2 + 27n + 3) - (16n^2 + 12n + 2)\}\) | dM1 |
| \(= \dfrac{n}{6}\{(38n^2 + 15n + 1)\}\) \((a = 38, b = 15, c = 1)\) | A1 |
| (4) | |
| (5 marks) |
Notes
B1: States or uses \(\displaystyle\sum_{r=1}^{2n} r^2 = \frac{1}{3}n(2n + 1)(4n + 1)\)
M1: Subtracts their sum to \(2n\) or \(2n - 1\) and attempts to factorise by \(\dfrac{n}{6}\) seen anywhere.
dM1: Expands two quadratics dependent on first M1
A1: cao