In expansion of {left( {frac{{3{x^2}}}{2} - frac{1}{{3x}}} right)^9} ,term independent of x is

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7.Binomial Theorem

easy

In the expansion of ${\left( {\frac{{3{x^2}}}{2} - \frac{1}{{3x}}} \right)^9}$,the term independent of $x$ is

$^9{C_3}.\frac{1}{{{6^3}}}$

$^9{C_3}{\left( {\frac{3}{2}} \right)^3}$

$^9{C_3}$

None of these

Solution

(a) In the expansion of ${\left( {\frac{{3{x^2}}}{2} + \frac{1}{{3x}}} \right)^9}$,

the general term is ${T_{r + 1}} = {\,^9}{C_r}.{\left( {\frac{{3{x^2}}}{2}} \right)^{9 – r}}{\left( { – \frac{1}{{3x}}} \right)^r}$

$ = {\,^9}{C_r}{\left( {\frac{3}{2}} \right)^{9 – r}}{\left( { – \frac{1}{3}} \right)^r}{x^{18 – 3r}}$

For the term independent of $x$, $18 -3r = 0$ ==> $ r = 6$

This gives the independent term${T_{6 + 1}} = {\,^9}{C_6}{\left( {\frac{3}{2}} \right)^{9 – 6}}{\left( { – \frac{1}{3}} \right)^6} = {\,^9}{C_3}.\frac{1}{{{6^3}}}$

Standard 11

Mathematics

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(IIT-2016)

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hard

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hard

(JEE MAIN-2019)

In the expansion of ${\left( {\frac{{3{x^2}}}{2} - \frac{1}{{3x}}} \right)^9}$,the term independent of $x$ is

Solution

Similar Questions

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