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Find the Coefficient Of: (Iv) X 9 in the Expansion of ( X 2 − 1 3 X ) 9 - Mathematics

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प्रश्न

Find the coefficient of: 

(iv)  \[x^9\]  in the expansion of  \[\left( x^2 - \frac{1}{3x} \right)^9\]

 

 

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उत्तर

(iv) Suppose x9 occurs at the (+ 1)th term in the above expression.

Then, we have:

\[T_{r + 1} = ^{9}{}{C}_r ( x^2 )^{9 - r} \left( \frac{- 1}{3x} \right)^r \]
\[ = ( - 1 )^r {9}{}{C}_r \left( x^{18 - 2r - r} \right) \left( \frac{1}{3^r} \right)\]
\[ \text{ For this term to contain } x^9 , \text{ we must have: } \]
\[18 - 3r = 9\]
\[ \Rightarrow 3r = 9\]
\[ \Rightarrow r = 3\]
\[ \therefore \text{ Coefficient of }  x^9 = ( - 1 )^3 {9}{}{C}_3 \frac{1}{3^3} = - \frac{9 \times 8 \times 7}{2 \times 9 \times 9} = \frac{- 28}{9}\]

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Introduction of Binomial Theorem
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 18: Binomial Theorem - Exercise 18.2 [पृष्ठ ३७]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 18 Binomial Theorem
Exercise 18.2 | Q 9.4 | पृष्ठ ३७

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