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Question
Find the middle terms in the expansion of:
(ii) \[\left( 2 x^2 - \frac{1}{x} \right)^7\]
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Solution
(ii) Here, n, i.e., 7, is an odd number.
\[Thus, the middle terms are \left( \frac{7 + 1}{2} \right)th and \left( \frac{7 + 1}{2} + 1 \right)th i . e . 4th and 5th\]
\[Now, \]
\[ T_4 = T_{3 + 1} \]
\[ = ^{7}{}{C}_3 (2 x^2 )^{7 - 3} \left( \frac{- 1}{x} \right)^3 \]
\[ = - \frac{7 \times 6 \times 5}{3 \times 2} \times 16 x^8 \times \frac{1}{x^3}\]
\[ = - 560 x^5 \]
\[\text{ And, } \]
\[ T_5 = T_{4 + 1} \]
\[ =^{7}{}{C}_4 (2 x^2 )^{7 - 4} \left( \frac{- 1}{x} \right)^4 \]
\[ = 35 \times 8 \times x^6 \times \frac{1}{x^4}\]
\[ = 280 x^2\]
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