In the expansion of {left( {frac{{3{x^2}}}{2} | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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{{

Vedclass · Accepted Answer

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

Vedclass · Answer

(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${T_{6 + 1}} = {\,^9}{C_6}{\left( {\frac{3}{2}} \right)^{9 - 6}}{\left( { - \frac{1}{3}} \right)^6} = {\,^9}{C_3}.\frac{1}{{{6^3}}}$

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

Similar Questions

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

The total number or irrational terms in the binomial expansion of $\left( {{7^{1/5}} - {3^{1/10}}} \right)^{60}$ is

The coefficients of three successive terms in the expansion of ${(1 + x)^n}$ are $165, 330$ and $462$ respectively, then the value of n will be

The number of positive integers $k$ such that the constant term in the binomial expansion of $\left(2 x^{3}+\frac{3}{x^{k}}\right)^{12}, x \neq 0$ is $2^{8} \cdot \ell$, where $\ell$ is an odd integer, is......

The term independent of $x$ in ${\left( {2x - \frac{1}{{2{x^2}}}} \right)^{12}}$is