| Date | November 2016 | Marks available | 4 | Reference code | 16N.1.hl.TZ0.6 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Prove that | Question number | 6 | Adapted from | N/A |
Question
The sum of the first \(n\) terms of a sequence \(\{ {u_n}\} \) is given by \({S_n} = 3{n^2} - 2n\), where \(n \in {\mathbb{Z}^ + }\).
Write down the value of \({u_1}\).
Find the value of \({u_6}\).
Prove that \(\{ {u_n}\} \) is an arithmetic sequence, stating clearly its common difference.
Markscheme
\({u_1} = 1\) A1
[1 mark]
\({u_6} = {S_6} - {S_5} = 31\) M1A1
[2 marks]
\({u_n} = {S_n} - {S_{n - 1}}\) M1
\( = (3{n^2} - 2n) - \left( {3{{(n - 1)}^2} - 2(n - 1)} \right)\)
\( = (3{n^2} - 2n) - (3{n^2} - 6n + 3 - 2n + 2)\)
\( = 6n - 5\) A1
\(d = {u_{n + 1}} - {u_n}\) R1
\( = 6n + 6 - 5 - 6n + 5\)
\( = \left( {6(n + 1) - 5} \right) - (6n - 5)\)
\( = 6\) (constant) A1
Notes: Award R1 only if candidate provides a clear argument that proves that the difference between ANY two consecutive terms of the sequence is constant. Do not accept examples involving particular terms of the sequence nor circular reasoning arguments (eg use of formulas of APs to prove that it is an AP). Last A1 is independent of R1.
[4 marks]
