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Date	May 2022	Marks available	2	Reference code	22M.1.SL.TZ2.6
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Show that	Question number	6	Adapted from	N/A

Question

Consider the binomial expansion ${(x + 1)}^{7} = x^{7} + a x^{6} + b x^{5} + 35 x^{4} + \dots + 1$ where $x \neq 0$ and $a, b \in ℤ^{+}$ .

Show that $b = 21$ .

[2]

The third term in the expansion is the mean of the second term and the fourth term in the expansion.

Find the possible values of $x$ .

[5]

Markscheme

EITHER

recognises the required term (or coefficient) in the expansion (M1)

$b x^{5} = {}_{7}C_{2} x^{5} 1^{2}$ OR $b = {}_{7}C_{2}$ OR ${}_{7}C_{5}$

$b = \frac{7!}{2! 5!} (= \frac{7!}{2! (7 - 2)!})$

correct working A1

$\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 5 \times 4 \times 3 \times 2 \times 1}$ OR $\frac{7 \times 6}{2!}$ OR $\frac{42}{2}$

lists terms from row $7$ of Pascal’s triangle (M1)

$1, 7, 21, \dots$ A1

THEN

$b = 21$ AG

[2 marks]

$a = 7$ (A1)

correct equation A1

$21 x^{5} = \frac{a x^{6} + 35 x^{4}}{2}$ OR $21 x^{5} = \frac{7 x^{6} + 35 x^{4}}{2}$

correct quadratic equation A1

$7 x^{2} - 42 x + 35 = 0$ OR $x^{2} - 6 x + 5 = 0$ (or equivalent)

valid attempt to solve their quadratic (M1)

$(x - 1) (x - 5) = 0$ OR $x = \frac{6 \pm \sqrt{{(- 6)}^{2} - 4 (1) (5)}}{2 (1)}$

$x = 1, x = 5$ A1

Note: Award final A0 for obtaining $x = 0, x = 1, x = 5$ .

[5 marks]

Examiners report

The majority of candidates answered part (a) correctly, either by using the ${}^{n}C_{r}$ formula or Pascal's Triangle. In part (b) of the question, most candidates were able to correctly find the value of $a = 7$ and set up a correct equation showing the mean of the second and fourth terms. While some struggled to complete the required algebra to solve the equation, the majority of candidates who found a correct quadratic equation were able to solve it correctly. A few candidates included $x = 0$ in their final answer, thus not earning the final mark.

[N/A]

Syllabus sections

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

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