FP1 January 2013 Q1
1. Show, using the formulae for \(\displaystyle\sum_{r=1}^{n} r\) and \(\displaystyle\sum_{r=1}^{n} r^2\), that \[\sum_{r=1}^{n} 3(2r - 1)^2 = n(2n + 1)(2n - 1), \text{ for all positive integers } n.\] (5)
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{n}3(4r^2 - 4r + 1) = 12\sum_{r=1}^{n}r^2 - 12\sum_{r=1}^{n}r + \sum_{r=1}^{n}3\) | M1 |
| \(= \dfrac{12}{6}n(n + 1)(2n + 1) - \dfrac{12}{2}n(n + 1),\quad +3n\) | A1, B1 |
| \(= n\left[2(n + 1)(2n + 1) - 6(n + 1) + 3\right]\) | M1 |
| \(= n\left[4n^2 - 1\right] = n(2n + 1)(2n - 1)\) | A1 cso |
| [5] |
Notes
Induction is not acceptable here
First M for expanding given expression to give a 3 term quadratic and attempt to substitute.
First A for first two terms correct or equivalent.
B for \(+3n\) appearing
Second M for factorising by \(n\)
Final A for completely correct solution