$(1+x{)}^n=1+nx+{\displaystyle \frac{n(n-1)}{2!}}{x}^2+{\displaystyle \frac{n(n-1)(n-2)}{3!}}{x}^3+\dots$

$n$ is rational and $|x|<1$

Example:

Expand $(2+x^2{)}^{-2}$ in ascending powers of $x$, up to and including the term of $x^4$, simplifying the coefficient. Find the set of values of $x$ for which the expansion is valid.

Expansion of $(2+x^2{)}^{-2}$

$$\begin{array}{rl}& {(2.[1+{\displaystyle \frac{1}{2}}{x}^2])}^{-2}\\ \\ & 2^{-2}.{[1+{\displaystyle \frac{1}{2}}{x}^2]}^{-2}\\ \\ & {\displaystyle \frac{1}{4}}.[1+(-2).{\displaystyle \frac{1}{2}}{x}^2+{\displaystyle \frac{(-2).(-2-1)}{2!}}{\left({\displaystyle \frac{1}{2}}{x}^2\right)}^2]\\ \\ & {\displaystyle \frac{1}{4}}.[1-x^2+{\displaystyle \frac{3}{4}}{x}^4]\\ \\ & {\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}{x}^2+{\displaystyle \frac{3}{16}}{x}^4\end{array}$$

The expansion is valid when $\left|{\displaystyle \frac{1}{2}}{x}^2\right|<1$

$$\begin{array}{rl}& \left|{\displaystyle \frac{1}{2}}{x}^2\right|<1\\ \\ & {\displaystyle \frac{1}{2}}\left|x^2\right|<1\\ \\ & \left|x^2\right|<2\\ \\ & x^2<2\\ \\ & x^2-2<0\\ \\ & (x+\sqrt{2})(x-\sqrt{2})<0\\ \\ & -\sqrt{2}<x<\sqrt{2}\end{array}$$