Question

Find the middle terms in the expansion of:

(iv)  \[\left( x^4 - \frac{1}{x^3} \right)^{11}\]

 

Answer 1

\[\text{ Here, n, i . e . , 11, is an odd number }  . \]
\[\text{ Thus, the middle terms are } \left( \frac{11 + 1}{2} \right)\text{ th and } \left( \frac{11 + 1}{2} + 1 \right)\text{ th i . e . 6th and 7th}  . \]
\[\text{ Now,}  \]
\[ T_6 = T_{5 + 1} \]
\[ = ^{11}{}{C}_5 ( x^4 )^{11 - 5} \left( \frac{- 1}{x^3} \right)^5 \]
\[ = - \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2} \times \left( x \right)^{24 - 15} \]
\[ = - 462 x^9 \]
\[\text{ And, } \]
\[ T_7 = T_{6 + 1} \]
\[ = ^ {11}{}{C}_6 ( x^4 )^{11 - 6} \left( \frac{- 1}{x^3} \right)^6 \]
\[ = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2} \left( x \right)^{20 - 18} \]
\[ = 462 x^2 \]