Kausani Dutta

Problem:The term that is independent of x in the expansion of

[\frac{3x^2}{2}-\frac{1}{3x}]^9

a) \binom{9}{6}(\frac{1}{3})^3(\frac{3}{2})^6

b) \binom{9}{5}(\frac{3}{2})^5(-\frac{1}{3})^4

c) \binom{9}{3}(\frac{1}{6})^3

d) \binom{9}{4}(\frac{3}{2})^4(-\frac{1}{3})^5

Solution:

[\frac{3x^2}{2}-\frac{1}{3}]^9  =(\frac{3x^2}{2})^9+\binom{9}{1}(\frac{3x^2}{2})^8(-\frac{1}{3x})+...\binom{9}{6}(\frac{3x^2}{2})^3(-\frac{1}{3x})^6+...-(\frac{1}{3x})^9

Therefore the term independent of x is

=\binom{9}{6}(\frac{3}{2})^3(\frac{1}{3})^6

=\frac{9!}{6!3!}[\frac{3x3x3}{2x2x2x3x3x3x3x3x3}]

=\binom{9}{3}(\frac{1}{6})^3 (c)