Question 312 of 347MediumMCQ
If the term independent of x in the expansion of
\[{\left( {{3 \over 2}{x^2} - {1 \over {3x}}} \right)^9}\] is k, then 18 k is equal to :
Solutions · 1
Official solution
General term,
\[{T_{r + 1}} = {}^9{C_r}{\left[ {{3 \over 2}{x^2}} \right]^{9 - r}}{\left( { - {1 \over {3x}}} \right)^r}\]
\[{T_{r + 1}} = {}^9{C_r}{\left[ {{3 \over 2}} \right]^{9 - r}}{\left( { - {1 \over 3}} \right)^r}{x^{18 - 3r}}\]
For independent of x
18 − 3r = 0 \[ \Rightarrow \] r = 6
\[ \therefore \] \[{T_7} = {}^9{C_6}{\left( {{3 \over 2}} \right)^3}{\left( { - {1 \over 3}} \right)^6} = {{21} \over {54}} = k\]
\[ \therefore \] \[18k = {{21} \over {54}} \times 18 = 7\] Answer the question to unlock