Question:
In the expansion of $\left(x^{2}-\frac{1}{x^{2}}\right)^{16}$, the value of the constant term is ________________
Solution:
$\operatorname{In}\left(x^{2}-\frac{1}{x^{2}}\right)^{16}$
$T_{r+1}={ }^{16} C_{r}\left(x^{2}\right)^{16-r}\left(\frac{-1}{x^{2}}\right)^{r}$
$={ }^{16} C_{r} x^{32-2 r}(-1)^{r} x^{-2 r}$
i.e. $T_{r+1}={ }^{16} C_{r}(-1)^{r} x^{32-4 r}$
for constant term, $x^{32-4 r}=x^{0}$
i.e. $32-4 r=0$
i.e. $r=8$
$\therefore T_{8+1}={ }^{16} C_{8}$
i.e value of constant term isĀ 16C8.