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)

${^{16}C_{8}}$

Answer 2

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, $T_9 = {^{16}C_{8}}$