This is a problem from Test of Mathematics, TOMATO Problem 106. It is useful for ISI and CMI Entrance Exam. Try out the problem.
Problem: TOMATO Objective 106
The term that is independent of x in the expansion of
$[\frac{3x^2}{2}-\frac{1}{3x}]^9 $
a) ${{9} \choose {6}}(\frac{1}{3})^3(\frac{3}{2})^6 $
b) ${{9} \choose {5}}(\frac{3}{2})^5(-\frac{1}{3})^4 $
c) ${{9} \choose {3}}(\frac{1}{6})^3 $
d) ${{9} \choose {4}}(\frac{3}{2})^4(-\frac{1}{3})^5 $
Solution:
$[\frac{3x^2}{2}-\frac{1}{3}]^9$
=$(\frac{3x^2}{2})^9+{{9}\choose{1}}(\frac{3x^2}{2})^8(-\frac{1}{3x})+...{{9}\choose{6}}(\frac{3x^2}{2})^3(-\frac{1}{3x})^6+...-(\frac{1}{3x})^9 $
Therefore the term independent of x is
=${{9}\choose{6}}(\frac{3}{2})^3(\frac{1}{3})^6 $
=$\frac{9!}{6!3!}[\frac{3x3x3}{2x2x2x3x3x3x3x3x3}] $
=${{9}\choose{3}}(\frac{1}{6})^3 $ (c)
This is a problem from Test of Mathematics, TOMATO Problem 106. It is useful for ISI and CMI Entrance Exam. Try out the problem.
Problem: TOMATO Objective 106
The term that is independent of x in the expansion of
$[\frac{3x^2}{2}-\frac{1}{3x}]^9 $
a) ${{9} \choose {6}}(\frac{1}{3})^3(\frac{3}{2})^6 $
b) ${{9} \choose {5}}(\frac{3}{2})^5(-\frac{1}{3})^4 $
c) ${{9} \choose {3}}(\frac{1}{6})^3 $
d) ${{9} \choose {4}}(\frac{3}{2})^4(-\frac{1}{3})^5 $
Solution:
$[\frac{3x^2}{2}-\frac{1}{3}]^9$
=$(\frac{3x^2}{2})^9+{{9}\choose{1}}(\frac{3x^2}{2})^8(-\frac{1}{3x})+...{{9}\choose{6}}(\frac{3x^2}{2})^3(-\frac{1}{3x})^6+...-(\frac{1}{3x})^9 $
Therefore the term independent of x is
=${{9}\choose{6}}(\frac{3}{2})^3(\frac{1}{3})^6 $
=$\frac{9!}{6!3!}[\frac{3x3x3}{2x2x2x3x3x3x3x3x3}] $
=${{9}\choose{3}}(\frac{1}{6})^3 $ (c)
