Question

In the expansion of ${\left(x^2-\frac{1}{x^2}\right)}^{16},$ the value of the constant term is ___________.

Answer 1

$$\begin{gathered} \mathrm{In}{\left(x^2-\frac{1}{x^2}\right)}^{16} \\ {T}_{r+1}={}^{16}C_r{\left(x^2\right)}^{16-r}{\left(\frac{-1}{x^2}\right)}^r \\ ={}^{16}C_r{x}^{32-2r}{\left(-1\right)}^r{x}^{-2r} \\ \mathrm{i}.\mathrm{e}.T_{r+1}={}^{16}C_r{\left(-1\right)}^r{x}^{32-4r} \\ \mathrm{for}\mathrm{constant}\mathrm{term},x^{32-4r}=x^0 \\ \mathrm{i}.\mathrm{e}.32-4r=0 \\ \mathrm{i}.\mathrm{e}.r=8 \\ \therefore T_{8+1}={}^{16}C_8 \end{gathered}$$
i.e value of constant term is ¹⁶C₈.