Question

The constant term in the expansion of ${\left(x^2-\frac{1}{x^2}\right)}^{16}$ is

• A. ${}^{16}{C}_8$
• B. ${}^{16}{C}_7$
• C. ${}^{16}{C}_9$
• D. ${}^{16}{C}_{10}$

Answer 1

Answer: (A)

General term, $T_{r+1}={}^{16}{C}_r(x^2{)}^{16-r}{\left(-\frac{1}{x^2}\right)}^r={}^{16}{C}_r(-1{)}^r{x}^{32-4r}$
For constant term, $32-4r=0$
$\Rightarrow$ r = 8
$\therefore$ Constant term, $T_9={}^{16}{C}_8$