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

KCETKCET 2018

Solution:

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