Question

The term independent of x in the expansion of $(1+x+2x^3){(\frac{3x^2}{2}-\frac{1}{3x})}^{9}is$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$In{(\frac{3x^2}{2}-\frac{1}{3x})}^9,T_{r+1}{=}^9{C}_r{\left(\frac{3x^2}{2}\right)}^{9-r}{\left(\frac{-1}{3x}\right)}^r{=}^9{C}_r{3}^{9-2r}(-1)r\frac{x^{18-2r-r}}{2^{9-r}}$

$18–2r–r=0\Rightarrow 18–3r=0\Rightarrow r=6$

$18–2r–r=\Rightarrow 18–3r=0\Rightarrow r=6$

18 – 2r – r = 18 – 3r = – 1 $\Rightarrow$ has no integral value.

$18–2r–r=–3\Rightarrow 18–3r=–3\Rightarrow 3r–21\Rightarrow r=7$

Required term $=\frac{{}^9{C}_6{3}^{-3}(-1{)}^6}{2^3+2}.\frac{{}^9{C}_7(-1{)}^7{3}^{-5}}{2^2}=\frac{84}{27\times 8}-\frac{2\times 36}{243\times 4}=\frac{7}{18}-\frac{2}{27}=\frac{17}{54}$