Question

Find the middle term in the expansion of ${\left(2ax-\frac{b}{x^2}\right)}^{12}$.

• A. $59136\frac{a^6{b}^6}{x^6}$
• B. $59130\frac{a^6{b}^6}{x^6}$
• C. $\frac{59132}{x^6}$
• D. $\frac{59130}{x^5}$

Answer 1

Answer: (A)

Since the power of the given expansion is even, it has one middle term which is ${\left(\frac{12+2}{2}\right)}^{th}=7^{th}$ term
$T_7={}^{12}{C}_6{\left(2ax\right)}^6{\left(\frac{-b}{x^2}\right)}^6$
$={}^{12}{C}_6\frac{2^6{a}^6{x}^6\cdot {\left(-b\right)}^6}{x^{12}}$
$={}^{12}{C}_6\frac{2^6{a}^6{b}^6}{x^6}$
$=\frac{59136a^6{b}^6}{x^6}$