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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)