AS June 2023 Paper 1 Q4
4 You are given that \(\displaystyle\sum_{r=1}^{n}(ar + b) = n^2\) for all \(n\), where \(a\) and \(b\) are constants.
By finding \(\displaystyle\sum_{r=1}^{n}(ar + b)\) in terms of \(a\), \(b\) and \(n\), determine the values of \(a\) and \(b\). [6]
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\sum_{r=1}^{n}(ar + b) = a\sum_{r=1}^{n} r + b\sum_{r=1}^{n} 1\) | M1 | 1.1a |
| \(\dfrac{1}{2}an(n + 1) + bn\) | A1 A1 | 1.2 1.1 |
| \(\dfrac{1}{2}an(n + 1) + bn = n^2\) \(\tfrac{1}{2}a = 1,\ \tfrac{1}{2}a + b = 0\) | M1 | 3.1a |
| So \(a = 2\) | A1 | 1.1 |
| \(b = -1\) | A1 | 1.1 |
| [6] |
Notes
M1: (1st) splitting sum
A1 A1: \(\dfrac{1}{2}an(n + 1)\ldots\)
\(\ldots + bn\)
M1: (2nd) equating coefficients (or substituting two values for \(n\))
Alternative solution
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{n}(ar + b)\) is an AP with 1st term \(a + b\), common diff \(a\) | M1 |
| so \(\displaystyle\sum_{r=1}^{n}(ar + b) = \dfrac{n}{2}[2(a + b) + (n - 1)a]\) | M1A1 |
| \(= \dfrac{n}{2}[2a + 2b + na - a] = \dfrac{1}{2}na + nb + \dfrac{1}{2}n^2a\) \(\tfrac{1}{2}a = 1,\ \tfrac{1}{2}a + b = 0\) | M1 |
| So \(a = 2\) | A1 |
| \(b = -1\) | A1 |
M1A1: use of sum formula (need not be simplified)
M1: (3rd) equating coeffs or substituting two values for \(n\)
Alternative solution
| Scheme | Marks |
|---|---|
| substitute \(n = 1\): \(a + b = 1\) substitute \(n = 2\): \(3a + 2b = 4\) solving simultaneously: \(2a + 2b = 2\) | M1 |
| so \(a = 2\) | A1 |
| \(b = -1\) | A1 |
As the question requires finding \(\displaystyle\sum_{r=1}^{n}(ar + b)\) in terms of \(a\), \(b\) and \(n\), allow a maximum of 3 marks for this method