Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The coefficient of the term independent of x in the expansion of <img alt="\left(1+x+2 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9" src="https://entrancecorner.oncodecogs.com/gif.la

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

\frac{1}{3}


Option: 2

\frac{19}{54}


Option: 3

\frac{17}{54}


Option: 4

\frac{1}{4}


Answers (1)

best_answer

The general term in the expansion of \left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9 is 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\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right) x^{18-3 r}         .....(i)

Now, 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         .....(ii)

= Sum of the coefficient of the terms x^0, x^{-1} and x^{-3} in  \left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9

For x^{0}  in (i) above, . For x^{-1}  in (i) above, there exists no value of r and hence no such term exists. For x^{-3} in (i) 18-3 r=-3 \Rightarrow r=7

 For term independent of x, in (ii) the coefficient 
\begin{aligned} & =1 \times{ }^9 C_6(-1)^6\left(\frac{3}{2}\right)^{9-6}\left(\frac{1}{3}\right)^6+2 \times{ }^9 C_7(-1)^7\left(\frac{3}{2}\right)^{9-7}\left(\frac{1}{3}\right)^7 \\ & =\frac{9.8 .7}{1.2 .3} \cdot \frac{3^3}{2^3} \cdot \frac{1}{3^6}+2 \frac{9.8}{1.2}(-1) \frac{3^2}{2^2} \cdot \frac{1}{3^7}=\frac{7}{18}-\frac{2}{27}=\frac{17}{54} \end{aligned}

Posted by

Ritika Kankaria

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question