20N.2.AHL.TZ0.H_4

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

use of Binomial expansion to find a term in either ${\left(\frac{1}{3x^2}-\frac{x}{2}\right)}^9, {\left(\frac{1}{3x^{{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}-\frac{x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{2}\right)}^9, {\left(\frac{1}{3}-\frac{x^3}{2}\right)}^9, {\left(\frac{1}{3x^3}-\frac{1}{2}\right)}^9$ or ${\left(2-3x^3\right)}^9$         (M1)(A1)

Note: Award M1 for a product of three terms including a binomial coefficient and powers of the two terms, and A1 for a correct expression of a term in the expansion.

finding the powers required to be $2$ and $7$         (M1)(A1)

constant term is ${}_{9}C_2\times {\left(\frac{1}{3}\right)}^2\times {\left(-\frac{1}{2}\right)}^7$         (M1)

Note: Ignore all $x$’s in student’s expression.

therefore term independent of $x$ is $-\frac{1}{32} \left(=-0.03125\right)$       A1

[6 marks]

[N/A]