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Date May 2018 Marks available 3 Reference code 18M.2.hl.TZ2.5
Level HL only Paper 2 Time zone TZ2
Command term Hence and Find Question number 5 Adapted from N/A

Question

Express the binomial coefficient (3n+13n2)(3n+13n2) as a polynomial in nn.

[3]
a.

Hence find the least value of nn for which (3n+13n2)>106(3n+13n2)>106.

[3]
b.

Markscheme

(3n+13n2)=(3n+1)!(3n2)!3!(3n+13n2)=(3n+1)!(3n2)!3!     (M1)

=(3n+1)3n(3n1)3!=(3n+1)3n(3n1)3!     A1

=92n312n=92n312n or equivalent     A1

[3 marks]

a.

attempt to solve =92n312n>106=92n312n>106     (M1)

n>60.57n>60.57     (A1)

Note: Allow equality.

n=61n=61     A1

[3 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 1 - Core: Algebra » 1.3 » Counting principles, including permutations and combinations.
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