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Question: Find the term independent of \(x\) in \({\left( {\frac{{3{x^2}}}{2} - \frac{1}{{3x}}} \right)^9}\) ....

Find the term independent of xx in (3x22−13x)9{\left( {\frac{{3{x^2}}}{2} - \frac{1}{{3x}}} \right)^9} .
(a)615 (b)718 (c)78 (d)49  \left( a \right)\frac{6}{{15}} \\\ \left( b \right)\frac{7}{{18}} \\\ \left( c \right)\frac{7}{8} \\\ \left( d \right)\frac{4}{9} \\\

Explanation

Solution

Hint- Use general term of binomial expansion Tr+1=nCr(X)n−r(Y)r{T_{r + 1}} = {}^n{C_r}{\left( X \right)^{n - r}}{\left( Y \right)^r} , where n is positive integer.
As we know,
(X+Y)n=nC0(X)n+nC1(X)n−1(Y)+...........+nCr(X)n−r(Y)r+.........nCn−1(X)(Y)n−1+nCn(Y)n{\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 Tr+1=nCr(X)n−r(Y)r{T_{r + 1}} = {}^n{C_r}{\left( X \right)^{n - r}}{\left( Y \right)^r}.
(X+Y)n=(3x22−13x)9{\left( {X + Y} \right)^n} = {\left( {\frac{{3{x^2}}}{2} - \frac{1}{{3x}}} \right)^9}, compare value of X, Y and n.
X=3x22X = \frac{{3{x^2}}}{2} ,Y=−13xY = \frac{{ - 1}}{{3x}} and n=9n = 9 .
Now, General term of this expansion (3x22−13x)9{\left( {\frac{{3{x^2}}}{2} - \frac{1}{{3x}}} \right)^9} is Tr+1=nCr(X)n−r(Y)r{T_{r + 1}} = {}^n{C_r}{\left( X \right)^{n - r}}{\left( Y \right)^r}.
⇒Tr+1=9Cr(3x22)9−r(−13x)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.
⇒Tr+1=9Cr(32)9−r(−13)r(x)18−2r(1x)r ⇒Tr+1=9Cr(32)9−r(−13)r(x)18−3r..........(2)  \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 ⇒r=6  18 - 3r = 0 \\\ \Rightarrow r = 6 \\\
Put the value of r in (2) equation.
⇒T7=9C6(32)3(−13)6(x)0 ⇒T7=9C6(18)(127).............(3)  \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 nCr=n!r!(n−r)!^n{C_r} = \frac{{n!}}{{r!\left( {n - r} \right)!}} .

9C6=9!6!(9−6)!=9!6!(3)! ⇒9C6=9×8×7×6!6!(3)!=9×8×76=84  {}^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 9C6{}^9{C_6} in (3) equation.
⇒T7=848×27 ⇒T7=718  \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.