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Date May 2018 Marks available 7 Reference code 18M.2.hl.TZ2.6
Level HL only Paper 2 Time zone TZ2
Command term Use and Prove that Question number 6 Adapted from N/A

Question

Use mathematical induction to prove that (1a)n>1na for {n:nZ+,n where 0 < a < 1.

Markscheme

Let {{\text{P}}_n} be the statement: {\left( {1 - a} \right)^n} > 1 - na for some {n \in {\mathbb{Z}^ + },\,n \geqslant 2} where 0 < a < 1 consider the case n = 2{\text{:}}\,\,\,{\left( {1 - a} \right)^2} = 1 - 2a + {a^2}     M1

> 1 - 2a because {a^2} < 0. Therefore {{\text{P}}_2} is true     R1

assume {{\text{P}}_n} is true for some n = k

{\left( {1 - a} \right)^k} > 1 - ka     M1

Note: Assumption of truth must be present. Following marks are not dependent on this M1.

EITHER

consider {\left( {1 - a} \right)^{k + 1}} = \left( {1 - a} \right){\left( {1 - a} \right)^k}     M1

> 1 - \left( {k + 1} \right)a + k{a^2}      A1

> 1 - \left( {k + 1} \right)a \Rightarrow {{\text{P}}_{k + 1}} is true (as k{a^2} > 0)     R1

OR

multiply both sides by \left( {1 - a} \right) (which is positive)      M1

{\left( {1 - a} \right)^{k + 1}} > \left( {1 - ka} \right)\left( {1 - a} \right)

{\left( {1 - a} \right)^{k + 1}} > 1 - \left( {k + 1} \right)a + k{a^2}     A1

{\left( {1 - a} \right)^{k + 1}} > 1 - \left( {k + 1} \right)a \Rightarrow {{\text{P}}_{k + 1}} is true (as k{a^2} > 0)     R1

THEN

{{\text{P}}_2} is true {{\text{P}}_k} is true  \Rightarrow {{\text{P}}_{k + 1}} is true so {{\text{P}}_n} true for all n > 2 (or equivalent)      R1

Note: Only award the last R1 if at least four of the previous marks are gained including the A1.

[7 marks]

Examiners report

[N/A]

Syllabus sections

Topic 1 - Core: Algebra » 1.4 » Proof by mathematical induction.
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