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Date May 2008 Marks available 5 Reference code 08M.2.hl.TZ1.1
Level HL only Paper 2 Time zone TZ1
Command term Determine Question number 1 Adapted from N/A

Question

Determine the first three terms in the expansion of (12x)5(1+x)7 in ascending powers of x.

Markscheme

METHOD 1

constant term: (50)(2x)0(70)x0=1     A1

term in x: (71)x+(51)(2x)=3x     (M1)A1

term in x2 : (72)x2+(52)(2x)2+(71)x(51)(2x)=9x2     M1A1     N3

[5 marks]

METHOD 2

(12x)5(1+x)7=(1+5(2x)+5×4(2x)22!+...)(1+7x+7×62x2+...)     M1M1

=(110x+40x2+...)(1+7x+21x2+)

=1+7x+21x210x70x2+40x2+

=13x9x2+     A1A1A1     N3

[5 marks]

Examiners report

Although the majority of the candidates understood the question and attempted it, excessive time was spent on actually expanding the expression without consideration of the binomial theorem. A fair amount of students confused “ascending order”, giving the last three instead of the first three terms.

Syllabus sections

Topic 1 - Core: Algebra » 1.3 » The binomial theorem: expansion of (a+b)n, nN .
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