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Question
Find the middle terms(s) in the expansion of:
(iv) \[\left( 2x - \frac{x^2}{4} \right)^9\]
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Solution
\[\left( 2x - \frac{x^2}{4} \right)^9 \]
\[\text{ Here, n is an odd number } . \]
\[\text{ Therefore, the middle terms are } \left( \frac{n + 1}{2} \right)\text{ th and } \left( \frac{n + 1}{2} + 1 \right)\text{ th, i . e . 5th and 6th terms . } \]
\[\text{ Now, we have} \]
\[ T_5 = T_{4 + 1} \]
\[ =^{9}{}{C}_4 (2x )^{9 - 4} \left( \frac{- x^2}{4} \right)^4 \]
\[ = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} \times 2^5 \frac{1}{4^4} x^{5 + 8} \]
\[ = \frac{63}{4} x^{13} \]
\[\text{ And} , \]
\[ T_6 = T_{5 + 1} \]
\[ = ^{9}{}{C}_5 (2x )^{9 - 5} \left( \frac{- x^2}{4} \right)^5 \]
\[ = - \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} \times 2^4 \frac{1}{4^5} x^{4 + 10} \]
\[ = - \frac{63}{32} x^{14}\]
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