C1 June 2011 Q5
5. A sequence \(a_1, a_2, a_3, \ldots\) is defined by\[\begin{aligned}a_1 &= k,\\ a_{n+1} &= 5a_n + 3, \qquad n \geqslant 1,\end{aligned}\]where \(k\) is a positive integer.
| Scheme | Marks |
|---|---|
| \(\left(a_2 =\right)\ 5k + 3\) | B1 |
| (1) |
Notes
(a) \(5k + 3\) must be seen in (a) to gain the mark
| Scheme | Marks |
|---|---|
| \(\left(a_3 =\right)\ 5(5k + 3) + 3\) | M1 |
| \(= 25k + 18 \qquad (*)\) | A1 cso |
| (2) |
Notes
(b) 1st M: Substitutes their \(a_2\) into \(5a_2 + 3\) - note the answer is given so working must be seen.
| Scheme | Marks |
|---|---|
| (i) \(a_4 = 5(25k + 18) + 3 \quad (= 125k + 93)\) | M1 |
| \(\displaystyle\sum_{r=1}^{4} a_r = k + (5k + 3) + (25k + 18) + (125k + 93)\) | M1 |
| \(= 156k + 114\) | A1 cao |
| (ii) \(= 6(26k + 19)\) (or explain each term is divisible by 6) | A1 ft |
| (4) | |
| (7 marks) |
Notes
(c) 1st M1: Substitutes their \(a_3\) into \(5a_3 + 3\) or uses \(125k + 93\)
2nd M1: for their sum \(k + a_2 + a_3 + a_4\) - must see evidence of four terms with plus signs and must not be sum of AP
1st A1: All correct so far
2nd A1ft: Limited ft – previous answer must be divisible by 6 (eg \(156k + 42\)). This is dependent on second M mark in (c)
Allow \(\dfrac{156k + 114}{6} = 26k + 19\) without explanation. No conclusion is needed.