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Question
Find the middle terms in the expansion of:
(i) \[\left( 3x - \frac{x^3}{6} \right)^9\]
Solution
(i) Here, n, i.e. 9, is an odd number.
Thus, the middle terms are \[\left( \frac{n + 1}{2} \right)\text{th and } \left( \frac{n + 1}{2} + 1 \right)th, \text{ i . e . 5th and 6th } \]
\[Now, \]
\[ T_5 = T_{4 + 1} = ^{9}{}{C}_4 (3x )^{9 - 4} \left( \frac{- x^3}{6} \right)^4 \]
\[ = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} \times 27 \times 9 \times \frac{1}{36 \times 36} x^{17} \]
\[ = \frac{189}{8} x^{17} \]
\[\text{ and } , \]
\[ T_6 = T_{5 + 1} \]
\[ =^{9}{}{C}_5 (3x )^{9 - 5} \left( \frac{- x^3}{6} \right)^5 \]
\[ = - \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} \times 81 \times \frac{1}{216 \times 36} x^{19} \]
\[ = - \frac{21}{16} x^{19}\]
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