Answer
$1-2x+\dfrac{3x^2}{2}-\dfrac{x^3}{2}+\dfrac{x^4}{16}$
Work Step by Step
The formula to determine the binomial series is:
$(1+x)^p=1+\Sigma_{k=1}^\infty \dbinom{p}{k}x^k$
Here, $\dbinom{p}{k}=\dfrac{p(p-1)(p-2).....(p-k+1)}{k!}$
Now, $(1-\dfrac{x}{2})^{3}=1+(4)(-\dfrac{x}{2})+\dfrac{(4)(3)(x^2/4)}{2!}+\dfrac{(4)(3)(2)(-x^3/8)}{3!}+\dfrac{(4)(3)(2)(1)(x^4/16)}{4!}=1-2x+\dfrac{3x^2}{2}-\dfrac{x^3}{2}+\dfrac{x^4}{16}$
