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:

$latex [\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)