## Thomas' Calculus 13th Edition

$1-2x+\dfrac{3}{2}x^2-\dfrac{1}{8}x^3+\dfrac{1}{16}x^4$
Apply the binomial series formula to determine the first four terms: $(1+x)^r=1+\Sigma_{k=1}^\infty \dbinom{r}{k}x^k$ Here, we have $\dbinom{r}{k}=\dfrac{r(r-1)(r-2).....(r-k+1)}{k!}$ $(1-\dfrac{x}{2})^{3}=1+(4)(-\dfrac{x}{2})+\dfrac{(4)(3)}{2!}(\dfrac{x^2}{4})+\dfrac{(4)(3)(2)}{3!}(\dfrac{-x^3}{8})+\dfrac{(4)(3)(2)(1)}{4!}(\dfrac{x^4}{16})=1-2x+\dfrac{3}{2}x^2-\dfrac{1}{8}x^3+\dfrac{1}{16}x^4$