Question

Find the term independent of $x$ in the expansion of following expression
${\left( {\dfrac{3}{2}{x^2} - \dfrac{1}{{3x}}} \right)^6}$

Answer 1

We know the general term of expansion of ${\left( {a + b} \right)^n}$ which is ${T_{r + 1}} = {}^n{C_r}{(a)^{n - r}}.{(b)^n}$ . we will compare the this equation with the equation given in problem. We need to find the term independent of x so we will take power x as 0. After comparing the equations we will get the value of r. After putting the value of r in the equation we will get an answer.

Complete step by step solution:
We know that general term of expansion ${\left( {a + b} \right)^n}$ is
${T_{r + 1}} = {}^n{C_r}{(a)^{n - r}}.{(b)^n}$
For general term of expansion ${\left( {\dfrac{3}{2}{x^2} - \dfrac{1}{{3x}}} \right)^6}$
Putting

& n = 6,a = \dfrac{3}{2}{x^2},b = - \dfrac{1}{{3x}} \cr & {T_{r + 1}} = {}^6{C_r}{(\dfrac{3}{2}{x^2})^{6 - r}}.{( - \dfrac{1}{{3x}})^r} \cr & = {}^6{C_r}{\left( {\dfrac{3}{2}} \right)^{6 - r}}{({x^2})^{6 - r}}.{\left( {\dfrac{{ - 1}}{3} \times \dfrac{1}{x}} \right)^r} \cr & = {}^6{C_r}{\left( {\dfrac{3}{2}} \right)^{6 - r}}{(x)^{2(6 - r)}}.{\left( {\dfrac{{ - 1}}{3}} \right)^r}{\left( {\dfrac{1}{x}} \right)^r} \cr & = {}^6{C_r}{\left( {\dfrac{3}{2}} \right)^{6 - r}}{(x)^{12 - 2r}}.{\left( {\dfrac{{ - 1}}{3}} \right)^r}{\left( x \right)^{ - r}} \cr & = {}^6{C_r}{\left( {\dfrac{3}{2}} \right)^{6 - r}}.{\left( {\dfrac{{ - 1}}{3}} \right)^r}{(x)^{12 - 2r}}{\left( x \right)^{ - r}} \cr & = {}^6{C_r}{\left( {\dfrac{3}{2}} \right)^{6 - r}}.{\left( {\dfrac{{ - 1}}{3}} \right)^r}{(x)^{12 - 2r - r}} \cr & = {}^6{C_r}{\left( {\dfrac{3}{2}} \right)^{6 - r}}.{\left( {\dfrac{{ - 1}}{3}} \right)^r}{(x)^{12 - 3r}} \cr} $$ We need to find the term independent of x So, power of x is 0 $${x^{12 - 3r}} = {x^0}$$ Comparing powers $$\eqalign{ \Rightarrow 12 - 3r = 0 \cr \Rightarrow 12 = 3r \cr \Rightarrow \dfrac{{12}}{3} = r \cr \Rightarrow 4 = r \cr \Rightarrow r = 4 \cr} $$ Putting r=4 in previous equation $$\eqalign{ & {T_{4 + 1}} = {}^6{C_r}{\left( {\dfrac{3}{2}} \right)^{6 - 4}}.{\left( {\dfrac{{ - 1}}{3}} \right)^4}{(x)^{12 - 3(4)}} \cr & {T_5} = {}^6{C_4}{\left( {\dfrac{3}{2}} \right)^2}.\left( {\dfrac{{ - 1}}{{{3^4}}}} \right){(x)^{12 - 12}} \cr & {T_5} = \dfrac{{6!}}{{4!(6 - 4)!}} \times \dfrac{1}{4} \times \dfrac{1}{9} \cr & = \dfrac{5}{{12}} \cr} $$ **Hence , the term which is independent of x is 5th term which is $$\dfrac{5}{{12}}$$** **Note:** We knew the general term of expansion of $${\left( {a + b} \right)^n}$$ which is $${T_{r + 1}} = {}^n{C_r}{(a)^{n - r}}.{(b)^n}$$ . here compare this equation with the equation given in the problem. And for the term independent of x so we will take power x as 0. After comparing the equations we will get the value of r. After putting the value of r in the equation we got the answer.