C1 June 2016 Q6
6. A sequence \(a_1, a_2, a_3, \ldots\) is defined by\[\begin{aligned} a_1 &= 4, \\ a_{n+1} &= 5 - ka_n, \quad n \geqslant 1 \end{aligned}\]where \(k\) is a constant.
Find
\(a_1 = 4,\ a_{n+1} = 5 - ka_n,\ n \geqslant 1\)
| Scheme | Marks |
|---|---|
| \(a_2 = 5 - ka_1 = 5 - 4k\) \(a_3 = 5 - ka_2 = 5 - k(5 - 4k)\) | M1A1 |
| (2) |
Notes
M1: Uses the recurrence relation correctly at least once. This may be implied by \(a_2 = 5 - 4k\) or by the use of \(a_3 = 5 - k\left(\text{their } a_2\right)\)
A1: Two correct expressions – need not be simplified but must be seen in (a).
Allow \(a_2 = 5 - k4\) and \(a_3 = 5 - 5k + k^2 4\)
Isw if necessary for \(a_3\).
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{3}(1) = 1 + 1 + 1\) | B1 |
| \(\displaystyle\sum_{r=1}^{3}a_r = 4 + \text{"}5 - 4k\text{"} + \text{"}5 - 5k + 4k^2\text{"}\) | M1 |
| \(\displaystyle\sum_{r=1}^{3}(1 + a_r) = 17 - 9k + 4k^2\) | A1 |
| (3) |
Notes
B1: Finds 1+1+1 or 3 somewhere in their solution (may be implied by e.g. \(5 + 6 - 4k + 6 - 5k + 4k^2\)). Note that \(5 + 6 - 4k + 6 - 5k + 4k^2\) would score B1 and the M1 below.
M1: Adds 4 to their \(a_2\) and their \(a_3\) where \(a_2\) and \(a_3\) are functions of \(k\). The statement as shown is sufficient.
A1: Cao but condone ‘= 0’ after the expression
Allow full marks in (b) for correct answer only
| Scheme | Marks |
|---|---|
| 500 | B1 |
| (1) | |
| (6 marks) |
Notes
B1: cao