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7.Binomial Theorem
medium
If the term independent of $x$ in the exapansion of $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$ is $k,$ then $18 k$ is equal to
A
$9$
B
$11$
C
$5$
D
$7$
(JEE MAIN-2020)
Solution
$T _{ r +1}={ }^{9} C _{ r }\left(\frac{3}{2} x ^{2}\right)^{9- r }\left(-\frac{1}{3 x }\right)^{ r }$
$T _{ r +1}={ }^{9} C _{ r }\left(\frac{3}{2}\right)^{9- r }\left(-\frac{1}{3}\right)^{ r } x ^{18-3 r }$
For independent of x
$18-3 r=0, r=6$
$\therefore \quad T _{7}={ }^{9} C _{6}\left(\frac{3}{2}\right)^{3}\left(-\frac{1}{3}\right)^{6}=\frac{21}{54}= k$
$\therefore \quad 18 k =\frac{21}{54} \times 18=7$
Standard 11
Mathematics
