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In the Expansion of ( X 2 − 1 3 X ) 9 , the Term Without X is Equal to - Mathematics

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Question

In the expansion of \[\left( x^2 - \frac{1}{3x} \right)^9\] , the term without x is equal to

 

Options

  •  \[\frac{28}{81}\]

  • \[\frac{-28}{243}\]

  • \[\frac{28}{243}\]

  •  none of these

     
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Solution

\[\frac{28}{243}\]

Suppose the (r + 1)th term in the given expansion is independent of x.
Then , we have:

\[T_{r + 1} = ^{9}{}{C}_r ( x^2 )^{9 - r} \left( \frac{- 1}{3x} \right)^r \]

`= ( - 1 )^r " ^9C_r \frac{1}{3^r} x^{18 - 2r - r}`

\[\text{ For this term to be independent of x, we must have: } \]

\[18 - 3r = 0\]

\[ \Rightarrow r = 6\]

`therefore \text{ Required term } = ( - 1 )^6 " ^9C_6 \frac{1}{3^6} = \frac{9 \times 8 \times 7}{3 \times 2} \times \frac{1}{3^6} = \frac{28}{243}`

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Chapter 18: Binomial Theorem - Exercise 18.4 [Page 46]

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RD Sharma Mathematics [English] Class 11
Chapter 18 Binomial Theorem
Exercise 18.4 | Q 8 | Page 46

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