## Thomas' Calculus 13th Edition

$1-x+\dfrac{3}{4}x^2-\dfrac{1}{2}x^3$
Apply Binomial series formula to find the first four terms. $(1+x)^m=1+\Sigma_{k=1}^\infty \dbinom{m}{k}x^k$ and $\dbinom{m}{k}=\dfrac{m(m-1)(m-2).....(m-k+1)}{k!}$ Thus, we have $(1+\dfrac{x}{2})^{-2}=1-(2)(\dfrac{x}{2})(-2x)+[\dfrac{(-2)(-3)(\dfrac{x}{2})^2}{2!}]+[\dfrac{(-2)(-3)(-4)(\dfrac{x}{2})^3}{3!}]+...$ or, $=1-x+\dfrac{(6)(\dfrac{1}{4})x^2}{2!}+\dfrac{(24)(\dfrac{1}{8})x^3}{3!}+...$ Hence, we have the first four terms: $1-x+\dfrac{3}{4}x^2-\dfrac{1}{2}x^3$