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Find the Middle Terms(S) in the Expansion Of: (Vii) ( 3 − X 3 6 ) 7 - Mathematics

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प्रश्न

Find the middle terms(s) in the expansion of: 

(vii) \[\left( 3 - \frac{x^3}{6} \right)^7\]

  

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उत्तर

\[\left( 3 - \frac{x^3}{6} \right)^7 \]
\[\text{ Here, n is an odd number } . \]
\[\text{ Therefore, the middle terms are }  \left( \frac{7 + 1}{2} \right)\text{ th and } \left( \frac{7 + 1}{2} + 1 \right)\text{ th, i . e . , 4th and 5th terms . } \]
\[\text{ Now, we have } \]
\[ T_4 = T_{3 + 1} \]
\[ =^{7}{}{C}_3 \left( 3 \right)^{7 - 3} \left( \frac{- x^3}{6} \right)^3 \]
\[ = - \frac{105}{8} x^9 \]
\[\text{ And} , \]
\[ T_5 = T_{4 + 1} \]
\[ = ^{9}{}{C}_4 \left( 3 \right)^{9 - 4} \left( \frac{- x^3}{6} \right)^4 \]
\[ = \frac{7 \times 6 \times 5}{3 \times 2} \times 3^5 \times \frac{1}{6^4} x^{12} \]
\[ = \frac{35}{48} x^{12}\]

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

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आरडी शर्मा Mathematics [English] Class 11
पाठ 18 Binomial Theorem
Exercise 18.2 | Q 15.07 | पृष्ठ ३८

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