Question

If the term independent of $x$ in the expansion 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

Answer 1

Answer: (D)

$T _{ r +1}={ }^{9} C _{ r }\left(\frac{3}{2} x ^{2}\right)^{9- r }\left(-\frac{1}{3 x }\right)^{ r }$
$T _{ r +1}={ }^{9} C _{ r }\left(\frac{3}{2}\right)^{9- r }\left(-\frac{1}{3}\right)^{ r } x ^{18-3 r }$
For independent of $x$
$ 18-3 r=0, r=6$
$\therefore T_{7}={ }^{9} C_{6}\left(\frac{3}{2}\right)^{3}\left(-\frac{1}{3}\right)^{6}=\frac{21}{54}=k $
$\therefore 18 k=\frac{21}{54} \times 18=7$