C1 January 2013 Q4
4. A sequence \(u_1, u_2, u_3, \ldots\) satisfies\[u_{n+1} = 2u_n - 1, \quad n \geqslant 1\]Given that \(u_2 = 9\),
(a) find the value of \(u_3\) and the value of \(u_4\), (2)
(b) evaluate \(\displaystyle\sum_{r=1}^{4} u_r\). (3)
| Scheme | Marks |
|---|---|
| \(u_2 = 9,\ u_{n+1} = 2u_n - 1,\ n \geqslant 1\) | |
| \(u_3 = 2u_2 - 1 = 2(9) - 1 \quad (= 17)\) \(u_3 = 2(9) - 1\). Can be implied by \(u_3 = 17\) | M1 |
| \(u_4 = 2u_3 - 1 = 2(17) - 1 = 33\) Both \(u_3 = 17\) and \(u_4 = 33\) | A1 |
| (2) |
Notes
M1: Substitutes 9 into RHS of iteration formula
A1: Needs both 17 and 33 (but allow if either or both seen in part (b))
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{4} u_r = u_1 + u_2 + u_3 + u_4\) | |
| \((u_1) = 5\) | B1 |
| \(\displaystyle\sum_{r=1}^{4} u_r = \text{"}5\text{"} + 9 + \text{"}17\text{"} + \text{"}33\text{"} = 64\) Adds their first four terms obtained legitimately (see notes below) | M1 |
| 64 | A1 |
| (3) | |
| (5 marks) |
Notes
B1: for \(u_1 = 5\) (however obtained – may appear in (a)) May be called \(a = 5\)
M1: Uses their \(u_1\) found from \(u_2 = 2u_1 - 1\) stated explicitly, or uses \(u_1 = 4\) or \(5\tfrac{1}{2}\), and adds it to \(u_2\), their \(u_3\) and their \(u_4\) only. (See special cases below).
There should be no fifth term included.
Use of sum of AP is irrelevant and scores M0
A1: 64