मराठी

Find the Middle Terms in the Expansion Of: (Ii) ( 2 X 2 − 1 X ) 7

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प्रश्न

Find the middle terms in the expansion of:

(ii) \[\left( 2 x^2 - \frac{1}{x} \right)^7\]

 

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उत्तर

(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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पाठ 18: Binomial Theorem - Exercise 18.2 [पृष्ठ ३८]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 18 Binomial Theorem
Exercise 18.2 | Q 14.2 | पृष्ठ ३८

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