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Date	May 2022	Marks available	5	Reference code	22M.1.AHL.TZ1.6
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Determine	Question number	6	Adapted from	N/A

Question

Consider the expansion of ${(8 x^{3} - \frac{1}{2 x})}^{n}$ where $n \in ℤ^{+}$ . Determine all possible values of $n$ for which the expansion has a non-zero constant term.

Markscheme

EITHER

attempt to obtain the general term of the expansion

$T_{r + 1} = {}_{n}C_{r} {(8 x^{3})}^{n - r} {(- \frac{1}{2 x})}^{r}$ OR $T_{r + 1} = {}_{n}C_{n - r} {(8 x^{3})}^{r} {(- \frac{1}{2 x})}^{n - r}$ (M1)

recognize power of $x$ starts at $3 n$ and goes down by $4$ each time (M1)

THEN

recognizing the constant term when the power of $x$ is zero (or equivalent) (M1)

$r = \frac{3 n}{4}$ or $n = \frac{4}{3} r$ or $3 n - 4 r = 0$ OR $3 r - (n - r) = 0$ (or equivalent) A1

$r$ is a multiple of $3 (r = 3, 6, 9, \dots)$ or one correct value of $n$ (seen anywhere) (A1)

$n = 4 k, k \in ℤ^{+}$ A1

Note: Accept $n$ is a (positive) multiple of $4$ or $n = 4, 8, 12, \dots$
Do not accept $n = 4, 8, 12$

Note: Award full marks for a correct answer using trial and error approach
showing $n = 4, 8, 12, \dots$ and for recognizing that this pattern continues.

[5 marks]

Examiners report

There was a mixed response to this question. Candidates who used a trial and error approach were more successful in obtaining completely correct answers than those who tried to solve algebraically by finding the general term to form an equation relating n and r . Poor explanations were often noted in the trial and error approach. Candidates often failed to make progress after obtaining $n = \frac{4}{3} r$ in the algebraic approach. Some candidates did not attempt this question.

Syllabus sections

Topic 1—Number and algebra » SL 1.9—Binomial theorem where n is an integer

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