Question

Find the term independent of $x$ in ${\left( {\frac{{3{x^2}}}{2} - \frac{1}{{3x}}} \right)^9}$ .
$
  \left( a \right)\frac{6}{{15}} \\
  \left( b \right)\frac{7}{{18}} \\
  \left( c \right)\frac{7}{8} \\
  \left( d \right)\frac{4}{9} \\
 $

Answer 1

Hint- Use general term of binomial expansion ${T_{r + 1}} = {}^n{C_r}{\left( X \right)^{n - r}}{\left( Y \right)^r}$ , where n is positive integer.
As we know,
${\left( {X + Y} \right)^n} = {}^n{C_0}{\left( X \right)^n}{ + ^n}{C_1}{\left( X \right)^{n - 1}}\left( Y \right) + ...........{ + ^n}{C_r}{\left( X \right)^{n - r}}{\left( Y \right)^r} + .........{}^n{C_{n - 1}}\left( X \right){\left( Y \right)^{n - 1}} + {}^n{C_n}{\left( Y \right)^n}$ is a binomial expansion, where n is positive integer and general term of this a binomial expansion is ${T_{r + 1}} = {}^n{C_r}{\left( X \right)^{n - r}}{\left( Y \right)^r}$.
${\left( {X + Y} \right)^n} = {\left( {\frac{{3{x^2}}}{2} - \frac{1}{{3x}}} \right)^9}$, compare value of X, Y and n.
$X = \frac{{3{x^2}}}{2}$ ,$Y = \frac{{ - 1}}{{3x}}$ and $n = 9$ .
Now, General term of this expansion ${\left( {\frac{{3{x^2}}}{2} - \frac{1}{{3x}}} \right)^9}$ is ${T_{r + 1}} = {}^n{C_r}{\left( X \right)^{n - r}}{\left( Y \right)^r}$.
$ \Rightarrow {T_{r + 1}} = {}^9{C_r}{\left( {\frac{{3{x^2}}}{2}} \right)^{9 - r}}{\left( {\frac{{ - 1}}{{3x}}} \right)^r}$
Now, collect all powers of x.
$
   \Rightarrow {T_{r + 1}} = {}^9{C_r}{\left( {\frac{3}{2}} \right)^{9 - r}}{\left( {\frac{{ - 1}}{3}} \right)^r}{\left( x \right)^{18 - 2r}}{\left( {\frac{1}{x}} \right)^r} \\
   \Rightarrow {T_{r + 1}} = {}^9{C_r}{\left( {\frac{3}{2}} \right)^{9 - r}}{\left( {\frac{{ - 1}}{3}} \right)^r}{\left( x \right)^{18 - 3r}}..........\left( 2 \right) \\
 $
To find the term independent of x. So, we have to make the power of x become 0.
$
  18 - 3r = 0 \\
   \Rightarrow r = 6 \\
$
Put the value of r in (2) equation.
$
   \Rightarrow {T_7} = {}^9{C_6}{\left( {\frac{3}{2}} \right)^3}{\left( {\frac{{ - 1}}{3}} \right)^6}{\left( x \right)^0} \\
   \Rightarrow {T_7} = {}^9{C_6}\left( {\frac{1}{8}} \right)\left( {\frac{1}{{27}}} \right).............\left( 3 \right) \\
 $
Now, we use $^n{C_r} = \frac{{n!}}{{r!\left( {n - r} \right)!}}$ .
\[
  {}^9{C_6} = \frac{{9!}}{{6!\left( {9 - 6} \right)!}} = \frac{{9!}}{{6!\left( 3 \right)!}} \\
   \Rightarrow {}^9{C_6} = \frac{{9 \times 8 \times 7 \times 6!}}{{6!\left( 3 \right)!}} = \frac{{9 \times 8 \times 7}}{6} = 84 \\
 \]
Put the value \[{}^9{C_6}\] in (3) equation.
$
   \Rightarrow {T_7} = \frac{{84}}{{8 \times 27}} \\
   \Rightarrow {T_7} = \frac{7}{{18}} \\
 $
So, the correct option is (b).
Note- Whenever we face such types of problems we use important points. Some points are to use the general term of binomial expansion and put the value of X and Y in the general term after comparison then make the power of x become zero for the independent term from x.