16N.1.AHL.TZ0.H_6

pestleMathematicsAIHLPaper 116N· sl-1-2-arithmetic-sequences-and-seriessource ↗

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 ${u_1}$ .

[1]

Find the value of ${u_6}$ .

[2]

Prove that $\{ {u_n}\}$ is an arithmetic sequence, stating clearly its common difference.

[4]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${u_1} = 1$ 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]

Examiners’ report

[N/A]