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Question
Find the middle terms(s) in the expansion of:
(ix) \[\left( \frac{p}{x} + \frac{x}{p} \right)^9\]
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Solution
\[\left( \frac{p}{x} + \frac{x}{p} \right)^9 \]
\[\text{ Here, n is an odd number } . \]
\[\text{ Therefore, the middle terms are } \left( \frac{9 + 1}{2} \right)^{th} \text{ and } \left( \frac{9 + 1}{2} + 1 \right)^{th} , i . e . , 5^{th} \text{ and } 6^{th}\text{ terms } . \]
\[\text{ Now, we have } \]
\[ T_5 = T_{4 + 1} \]
\[ = ^{9}{}{C}_4 \left( \frac{p}{x} \right)^{9 - 4} \left( \frac{x}{p} \right)^4 \]
\[ = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times \left( \frac{p}{x} \right)\]
\[ = \frac{126 p}{x}\]
\[\text{ And,} \]
\[ T_6 = T_{5 + 1} \]
\[ =^{9}{}{C}_5 \left( \frac{p}{x} \right)^{9 - 5} \left( \frac{x}{p} \right)^5 \]
\[ = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times \left( \frac{x}{p} \right)\]
\[ = \frac{126 x}{p}\]
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