FP1 January 2009 Q2
2.
(a) Show, using the formulae for \(\sum r\) and \(\sum r^2\), that \[\sum_{r=1}^{n}(6r^2 + 4r - 1) = n(n + 2)(2n + 1)\] (5)
(b) Hence, or otherwise, find the value of \(\displaystyle\sum_{r=11}^{20}(6r^2 + 4r - 1)\). (2)
| Scheme | Marks |
|---|---|
| \(6\sum r^2 + 4\sum r - \sum 1 = 6\dfrac{n}{6}(n + 1)(2n + 1) + 4\dfrac{n}{2}(n + 1),\ -n\) | M1 A1, B1 |
| \(= \dfrac{n}{6}(12n^2 + 18n + 6 + 12n + 12 - 6)\) or \(n(n + 1)(2n + 1) + (2n + 1)n\) | M1 |
| \(= \dfrac{n}{6}(12n^2 + 30n + 12) = n(2n^2 + 5n + 2) = n(n + 2)(2n + 1)\) * | A1 |
| (5) |
Notes
(a) First M1 for first 2 terms, B1 for \(-n\)
Second M1 for attempt to expand and gather terms.
Final A1 for correct solution only
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{20}(6r^2 + 4r - 1) - \sum_{r=1}^{10}(6r^2 + 4r - 1) = 20 \times 22 \times 41 - 10 \times 12 \times 21\) | M1 |
| \(= 15520\) | A1 |
| (2) | |
| [7] |
Notes
(b) Require (\(r\) from 1 to 20) subtract (\(r\) from 1 to 10) and attempt to substitute for M1