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  • If the term independent of  in the expansion of <img alt="\left ( \frac{3}{2}x^{2}-\frac{1}{3x} \right )^{9}" src="http://entrancecorner.codecogs.com/gif.latex?%5Cleft%2

If the term independent of x in the expansion of \left ( \frac{3}{2}x^{2}-\frac{1}{3x} \right )^{9} is k, then 18\; k is equal to :  
Option: 1 5
Option: 2 9
Option: 3 7
Option: 4 11

Answers (1)

best_answer

\mathrm{T}_{\mathrm{r}+1}={ }^{9} \mathrm{C}_{\mathrm{r}}\left(\frac{3}{2} \mathrm{x}^{2}\right)^{9-\mathrm{r}}\left(-\frac{1}{3 \mathrm{x}}\right)^{\mathrm{r}}

\mathrm{T}_{\mathrm{r}+1}={ }^{9} \mathrm{C}_{\mathrm{r}}\left(\frac{3}{2}\right)^{9-\mathrm{r}}\left(-\frac{1}{3}\right)^{\mathrm{r}} \mathrm{x}^{18-3 \mathrm{r}}

For independent of x

18 - 3r = 0

r = 6

\begin{aligned} \therefore \quad \mathrm{T}_{7}={ }^{9} \mathrm{C}_{6}\left(\frac{3}{2}\right)^{3}\left(-\frac{1}{3}\right)^{6}=\frac{21}{54}=\mathrm{k} \\ \therefore \quad 18 \mathrm{k}=\frac{21}{54} \times 18=7 \end{aligned}

Posted by

himanshu.meshram

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