Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 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) $\left( 3x - \frac{2}{x^2} \right)^{15}$

#### Solution

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

Is there an error in this question or solution?

#### APPEARS IN

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