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

Hint:
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
$$\eqalign{
  & n = 6,a = \dfrac{3}{2}{x^2},b = - \dfrac{1}{{3x}} \cr
  & {T_{r + 1}} = {}^6{C_r}{(\dfrac{3}{2}{x^2})^{6 - r}}.{( - \dfrac{1}{{3x}})^r} \cr
  & = {}^6{C_r}{\left( {\dfrac{3}{2}} \right)^{6 - r}}{({x^2})^{6 - r}}.{\left( {\dfrac{{ - 1}}{3} \times \dfrac{1}{x}} \right)^r} \cr
  & = {}^6{C_r}{\left( {\dfrac{3}{2}} \right)^{6 - r}}{(x)^{2(6 - r)}}.{\left( {\dfrac{{ - 1}}{3}} \right)^r}{\left( {\dfrac{1}{x}} \right)^r} \cr
  & = {}^6{C_r}{\left( {\dfrac{3}{2}} \right)^{6 - r}}{(x)^{12 - 2r}}.{\left( {\dfrac{{ - 1}}{3}} \right)^r}{\left( x \right)^{ - r}} \cr
  & = {}^6{C_r}{\left( {\dfrac{3}{2}} \right)^{6 - r}}.{\left( {\dfrac{{ - 1}}{3}} \right)^r}{(x)^{12 - 2r}}{\left( x \right)^{ - r}} \cr
  & = {}^6{C_r}{\left( {\dfrac{3}{2}} \right)^{6 - r}}.{\left( {\dfrac{{ - 1}}{3}} \right)^r}{(x)^{12 - 2r - r}} \cr
  & = {}^6{C_r}{\left( {\dfrac{3}{2}} \right)^{6 - r}}.{\left( {\dfrac{{ - 1}}{3}} \right)^r}{(x)^{12 - 3r}} \cr} $$
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.