Question

The middle term in the expansion of ${(ax-\frac{b}{x^2})}^{12}$ is:

• A. $\frac{984a^6{b}^6}{x^6}$
• B. $\frac{924a^5{b}^5}{x^5}$
• C. $\frac{628a^6{b}^6}{x^6}$
• D. $\frac{924a^6{b}^6}{x^6}$

Answer 1

The correct option is D $\frac{924a^6{b}^6}{x^6}$

Total number of the terms in the given expansion will be $n+1=13$
Hence, middle term of the expansion will be $\frac{13+1}{2}=7$

$\Rightarrow T_{6+1}={}^{12}{C}_6\times a^6{b}^6\times x^{-6}$
$=(\frac{12\times 11\times 10\times 9\times 8\times 7}{6\times 5\times 4\times 3\times 2\times 1}\times \frac{a^6{b}^6}{x^6})=\frac{924a^6{b}^6}{x^6}$

Hence, the middle term in the given expansion is $\frac{924a^6{b}^6}{x^6}.$