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Date November 2013 Marks available 7 Reference code 13N.1.hl.TZ0.6
Level HL only Paper 1 Time zone TZ0
Command term Prove Question number 6 Adapted from N/A

Question

Prove by mathematical induction that n3+11n is divisible by 3 for all nZ+.

Markscheme

n=1: 13+11=12

=3×4 or a multiple of 3     A1

assume the proposition is true for n=k   (ie k3+11k=3 m)     M1

 

Note:     Do not award M1 for statements with “Let n=k.

 

consider n=k+1:   (k+1)3+11(k+1)     M1

=k3+3k2+3k+1+11k+11     A1

=k3+11k+(3k2+3k+12)     M1

=3(m+k2+k+4)     A1

 

Note:     Accept k3+11k+3(k2+k+4) or statement that k3+11k+(3k2+3k+12) is a multiple of 3.

 

true for n=1, and n=k true n=k+1 true

hence true for all nZ+     R1

 

Note:     Only award the final R1 if at least 4 of the previous marks have been achieved.

 

[7 marks]

Examiners report

It was pleasing to see a great many clear and comprehensive answers for this relatively straightforward induction question. The inductive step only seemed to pose problems for the very weakest candidates. As in previous sessions, marks were mainly lost by candidates writing variations on ‘Let n=k’, rather than ‘Assume true for n=k’. The final reasoning step still needs attention, with variations on ‘n=k+1 true n=k true’ evident, suggesting that mathematical induction as a technique is not clearly understood.

Syllabus sections

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

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