Answer
$1-x+\dfrac{3}{4}x^2-\dfrac{1}{2}x^3+.....$
Work Step by Step
Formula to find the binomial series is:
$(1+x)^m=1+\Sigma_{k=1}^\infty \dbinom{m}{k}x^k$
Here, $\dbinom{m}{k}=\dfrac{m(m-1)(m-2).....(m-k+1)}{k!}$
Now, $(1+\dfrac{x}{2})^{-2}=1-2(\dfrac{x}{2})(-2x)+\dfrac{(-2)(-3)(\dfrac{x}{2})^2}{2!}+\dfrac{(-2)(-3)(-4)(\dfrac{x}{2})^3}{3!}+...$
or, $=1-x+\dfrac{(-2)(-3)(\dfrac{1}{4})x^2}{2!}+\dfrac{(-2)(-3)(-4)(\dfrac{1}{8})x^3}{3!}+...$
Thus, the first four terms are: $1-x+\dfrac{3}{4}x^2-\dfrac{1}{2}x^3+.....$
