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Find the term independent of x in the expansion of the expression: (i) ( 3 2 x 2 − 1 3 x ) 9

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Question

Find the term independent of x in the expansion of the expression: 

(i) \[\left( \frac{3}{2} x^2 - \frac{1}{3x} \right)^9\]

 

Short/Brief Note
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Solution

(i) Suppose the (r + 1)th term in the given expression is independent of x.
Now,

\[\left( \frac{3}{2} x^2 - \frac{1}{3x} \right)^9 \]
\[ T_{r + 1} =^{9}{}{C}_r \left( \frac{3}{2} x^2 \right)^{9 - r} \left( \frac{- 1}{3x} \right)^r \]
\[ = ( - 1 )^r  {9}{}{C}_r . \frac{3^{9 - 2r}}{2^{9 - r}} \times x^{18 - 2r - r} \]
\[\text{ For this term to be independent of x, we must have} \]
\[18 - 3r = 0\]
\[ \Rightarrow 3r = 18\]
\[ \Rightarrow r = 6\]
\[\text{ Hence, the required term is the 7th term } . \]
\[\text{ Now, we have } \]
\[ ^{9}{}{C}_6 \times \frac{3^{9 - 12}}{2^{9 - 6}}\]
\[ = \frac{9 \times 8 \times 7}{3 \times 2} \times 3^{- 3} \times 2^{- 3} \]
\[ = \frac{7}{18}\]

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Chapter 18: Binomial Theorem - Exercise 18.2 [Page 39]

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R.D. Sharma Mathematics [English] Class 11
Chapter 18 Binomial Theorem
Exercise 18.2 | Q 16.01 | Page 39

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