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

Option: 1

1/3    


Option: 2

19/54
   


Option: 3

17/54 


Option: 4

1/4


Answers (1)

best_answer

The (r + 1)th term in the expansion of [(3/2)x2 − (1/3x)]9 is given by
T_{r+1}={ }^9 C_r\left(\frac{3}{2} x^2\right)^{9-r}\left(-\frac{1}{3 x}\right)^r={ }^9 C_r(-1)^r \frac{3^{9-2 r}}{2^{9-r}} x^{18-3 r}      (1)

Since we are looking for the coefficient of the term independent of x in the expansion of
\left(1+x+2 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9            (2)

we must get the coefficient of x^0, x^{-1} and x^{-3} in the expansion of [(3/2)x^2-(1/3x)]^9. For x^0, r must be 6 in (1); for x^{-1}, there is no value of r ; and for x^{-3}, r must be 7 in (1). Therefore, the coefficient of the term independent of x in (2) is
\begin{aligned} &1 .{ }^9 C_6(-1)^6 \cdot \cdot^{\frac{3^{9-12}}{2^{9-6}}}+2 .{ }^9 C_7(-1)^7 \cdot \cdot^{\frac{3^{9-14}}{2^{9-7}}}\\ &=\frac{9.8 .7}{1.2 .3} \cdot \frac{3^{-3}}{2^3}+2 \cdot \frac{9.8}{1.2}(-1) \cdot \frac{3^{-5}}{2^2}=\frac{7}{18}-\frac{2}{27}=\frac{17}{54} \end{aligned}

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