C1 June 2007 Q8
8. A sequence \(a_1, a_2, a_3, \ldots\) is defined by\[\begin{aligned} a_1 &= k, \\ a_{n+1} &= 3a_n + 5, \qquad n \geqslant 1, \end{aligned}\]where \(k\) is a positive integer.
| Scheme | Marks |
|---|---|
| \(\left(a_2 =\right)\underline{3k + 5}\) [must be seen in part (a) or labelled \(a_2 =\) ] | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\left(a_3 =\right)3(3k + 5) + 5\) | M1 |
| \(= \underline{9k + 20} \qquad (*)\) | A1cso |
| (2) |
Notes
M1: for attempting to find \(a_3\), follow through their \(a_2 \neq k\).
A1cso: for simplifying to printed result with no incorrect working seen.
| Scheme | Marks |
|---|---|
| (i) \(a_4 = 3(9k + 20) + 5 \quad (= 27k + 65)\) | M1 |
| \(\displaystyle\sum_{r=1}^{4} a_r = k + (3k + 5) + (9k + 20) + (27k + 65)\) | M1 |
| (ii) \(= 40k + 90\) | A1 |
| \(= \underline{10(4k + 9)}\) (or explain why divisible by 10) | A1ft |
| (4) | |
| (7 marks) |
Notes
1st M1: for attempting to find \(a_4\). Can allow a slip here e.g. \(3(9k + 20)\) [i.e. forgot +5]
2nd M1: for attempting sum of 4 relevant terms, follow through their (a) and (b).
Must have 4 terms starting with \(k\).
Use of arithmetic series formulae at this point is M0A0A0
1st A1: for simplifying to \(40k + 90\) or better
2nd A1ft: for taking out a factor of 10 or dividing by 10 or an explanation in words true \(\forall k\).
Follow through their sum of 4 terms provided that both Ms are scored and their sum is divisible by 10.
A comment is not required.
e.g. \(\dfrac{40k + 90}{10} = 4k + 9\) is OK for this final A1.
S.C.: \(\displaystyle\sum_{r=2}^{5} a_r = 120k + 290 = 10(12k + 29)\) can have M1M0A0A1ft.