19M.2.AHL.TZ2.H_7

pestleMathematicsAIHLPaper 219M· sl-1-2-arithmetic-sequences-and-seriessource ↗

Suppose that ${u_1}$ is the first term of a geometric series with common ratio $r$ .

Prove, by mathematical induction, that the sum of the first $n$ terms, ${s_n}$ is given by

${s_n} = \frac{{{u_1}\left( {1 - {r^n}} \right)}}{{1 - r}}$ , where $n \in {\mathbb{Z}^ + }$ .

Markscheme / solution

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

$n = 1 \Rightarrow {s_1} = {u_1}$ , so true for $n = 1$ R1

assume true for $n = k$ , ie. ${s_k} = \frac{{{u_1}\left( {1 - {r^k}} \right)}}{{1 - r}}$ M1

Note: Award M0 for statements such as “let $n = k$ ”.

Note: Subsequent marks after the first M1 are independent of this mark and can be awarded.

${s_{k + 1}} = {s_k} + {u_1}{r^k}$ M1

${s_{k + 1}} = \frac{{{u_1}\left( {1 - {r^k}} \right)}}{{1 - r}} + {u_1}{r^k}$ A1

${s_{k + 1}} = \frac{{{u_1}\left( {1 - {r^k}} \right)}}{{1 - r}} + \frac{{{u_1}{r^k}\left( {1 - r} \right)}}{{1 - r}}$

${s_{k + 1}} = \frac{{{u_1} - {u_1}{r^k} + {u_1}{r^k} - r{u_1}{r^k}}}{{1 - r}}$ A1

${s_{k + 1}} = \frac{{{u_1}\left( {1 - {r^{k + 1}}} \right)}}{{1 - r}}$ A1

true for $n = 1$ and if true for $n = k$ then true for $n = k + 1$ , the statement is true for any positive integer (or equivalent). R1

Note: Award the final R1 mark provided at least four of the previous marks are gained.

[7 marks]

Examiners’ report

[N/A]