17M.2.AHL.TZ1.H_3

pestleMathematicsAAHLPaper 217M· sl-1-9-binomial-theorem-where-n-is-an-integersource ↗

The coefficient of ${x^2}$ in the expansion of ${\left( {\frac{1}{x} + 5x} \right)^8}$ is equal to the coefficient of ${x^4}$ in the expansion of ${\left( {a + 5x} \right)^7},{\text{ }}a \in \mathbb{R}$ . Find the value of $a$ .

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

$^8{C_r}{\left( {\frac{1}{x}} \right)^{8 - r}}{(5x)^r} = {}^8Cr{\left( 5 \right)^r}{x^{2r - 8}}$ (M1)

$r = 5$ (A1)

$^8{C_5} \times {5^5}{ = ^7}{C_4}{a^3} \times {5^4}$ M1A1

$56 \times 5 = 35{a^3}$

${a^3} = 8$ (A1)

$a = 2$ A1

METHOD 2

attempt to expand both binomials M1

$175000{x^2}$ A1

$21875{a^3}{x^4}$ A1

$175000 = 21875{a^3}$ M1

${a^3} = 8$ (A1)

$a = 2$ A1

[6 marks]

Examiners’ report

[N/A]