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Date May 2016 Marks available 1 Reference code 16M.1.hl.TZ0.6
Level HL only Paper 1 Time zone TZ0
Command term Conjecture Question number 6 Adapted from N/A

Question

Consider the recurrence relation Hn+1=2Hn+1, nZ+ where H1=1.

Find H2, H3 and H4.

[2]
a.

Conjecture a formula for Hn in terms of n, for nZ+.

[1]
b.

Prove by mathematical induction that your formula is valid for all nZ+.

[5]
c.

Markscheme

H2=2H1+1    (M1)

=3; H3=7; H4=15    A1

[2 marks]

a.

Hn=2n1    A1

[1 mark]

b.

let P(n) be the proposition that Hn=2n1 for nZ+

from (a) H1=1=211     A1

so P(1) is true

assume P(k) is true for some kHk=2k1     M1

then Hk+1=2Hk+1     (M1)

=2×(2k1)+1    A1

=2k+11

P(1) is true and P(k) is true P(k+1) is true, hence P(n) is true for all nZ+ by the principle of mathematical induction     R1

 

Note:     Only award the R1 if all earlier marks have been awarded.

 

[5 marks]

c.

Examiners report

This was a successful question for many candidates with many wholly correct answers seen. The vast majority of candidates were able to answer parts (a) and (b) and most knew how to start part (c). A few candidates were let down by not realising the need for a degree of formality in the presentation of the inductive proof.

a.

This was a successful question for many candidates with many wholly correct answers seen. The vast majority of candidates were able to answer parts (a) and (b) and most knew how to start part (c). A few candidates were let down by not realising the need for a degree of formality in the presentation of the inductive proof.

b.

This was a successful question for many candidates with many wholly correct answers seen. The vast majority of candidates were able to answer parts (a) and (b) and most knew how to start part (c). A few candidates were let down by not realising the need for a degree of formality in the presentation of the inductive proof.

c.

Syllabus sections

Topic 6 - Discrete mathematics » 6.11 » Recurrence relations. Initial conditions, recursive definition of a sequence.

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