In the expansion of

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.

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