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Date May 2016 Marks available 2 Reference code 16M.2.sl.TZ1.4
Level SL only Paper 2 Time zone TZ1
Command term Hence Question number 4 Adapted from N/A

Question

Find the term in x6 in the expansion of (x+2)9.

[4]
a.

Hence, find the term in x7 in the expansion of 5x(x+2)9.

[2]
b.

Markscheme

valid approach to find the required term     (M1)

eg(9r)(x)9r(2)r, x9+9x8(2)+(92)x7(2)2+, Pascal’s triangle to the 9th row

identifying correct term (may be indicated in expansion)     (A1)

eg4th term, r=6, (93), (x)6(2)3

correct calculation (may be seen in expansion)     (A1)

eg(93)(x)6(2)3, 84×23  

672x6   A1     N3

[4 marks]

a.

valid approach     (M1)

egrecognizing x7 is found when multiplying 5x×672x6

3360x7     A1     N2

[2 marks]

b.

Examiners report

Many candidates approached this question using an appropriate and efficient method to identify the required term. While many of those who were successful in (a) were also successful in (b), a significant number of candidates did not realize that multiplying their answer in (a) by 5x would give them the term in x7. This led to an attempt to find a binomial expansion, which was generally unsuccessful. Some candidates continue to be unable to distinguish between the “coefficient” and the “term” and lost a point as a result.

a.

Many candidates approached this question using an appropriate and efficient method to identify the required term. While many of those who were successful in (a) were also successful in (b), a significant number of candidates did not realize that multiplying their answer in (a) by 5x would give them the term in x7. This led to an attempt to find a binomial expansion, which was generally unsuccessful. Some candidates continue to be unable to distinguish between the “coefficient” and the “term” and lost a point as a result.

b.

Syllabus sections

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