Find the coefficient of:
(iv) \[x^9\] in the expansion of \[\left( x^2 - \frac{1}{3x} \right)^9\]
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Solution
(iv) Suppose x9 occurs at the (r + 1)th term in the above expression.
Then, we have:
\[T_{r + 1} = ^{9}{}{C}_r ( x^2 )^{9 - r} \left( \frac{- 1}{3x} \right)^r \]
\[ = ( - 1 )^r {9}{}{C}_r \left( x^{18 - 2r - r} \right) \left( \frac{1}{3^r} \right)\]
\[ \text{ For this term to contain } x^9 , \text{ we must have: } \]
\[18 - 3r = 9\]
\[ \Rightarrow 3r = 9\]
\[ \Rightarrow r = 3\]
\[ \therefore \text{ Coefficient of } x^9 = ( - 1 )^3 {9}{}{C}_3 \frac{1}{3^3} = - \frac{9 \times 8 \times 7}{2 \times 9 \times 9} = \frac{- 28}{9}\]
Concept: Introduction of Binomial Theorem
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