Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Find the Middle Terms in the Expansion Of: (Iii) ( 3 X − 2 X 2 ) 15 - Mathematics

Find the middle terms in the expansion of: 

(iii) \[\left( 3x - \frac{2}{x^2} \right)^{15}\]

 

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Solution

\[\text{ Given: } \]
\[\text{ n, i . e . 15 is an odd number}  . \]
\[\text{ Thus, the middle terms are}  \left( \frac{15 + 1}{2} \right)\text{ th and } \left( \frac{15 + 1}{2} + 1 \right)th \text{ i . e . 8th and 9th .}  \]
\[\text{ Now } , \]
\[ T_8 = T_{7 + 1} \]
\[ =^{15}{}{C}_7 (3x )^{15 - 7} \left( \frac{- 2}{x^2} \right)^7 \]
\[ = - \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9}{7 \times 6 \times 5 \times 4 \times 3 \times 2} \times 3^8 \times 2^7 x^{8 - 14} \]
\[ = \frac{- 6435 \times 3^8 \times 2^7}{x^6}\]
\[\text{ And } , \]
\[ T_9 = T_{8 + 1} \]
\[ =^{15}{}{C}_8 (3x )^{15 - 8} \left( \frac{- 2}{x^2} \right)^8 \]
\[ = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9}{7 \times 6 \times 5 \times 4 \times 3 \times 2} \times 3^7 \times 2^8 \times x^{7 - 16} \]
\[ = \frac{6435 \times 3^7 \times 2^8}{x^9}\]

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 18 Binomial Theorem
Exercise 18.2 | Q 14.3 | Page 38
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