Answer
$1-x-\dfrac{1}{2}x^2-\dfrac{1}{2}x^3$
Work Step by Step
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,$(1-2x)^{1/2}=1+\dfrac{1}{2}(-2x)+\dfrac{(\dfrac{1}{2})(-\dfrac{1}{2})(-2x)^2}{2!}+\dfrac{(\dfrac{1}{2})(-\dfrac{1}{2})(-\dfrac{3}{2})(-2x)^3}{3!}+...=1-x-\dfrac{1}{2}x^2-\dfrac{1}{2}x^3...$
Hence, we have the first four terms: $1-x-\dfrac{1}{2}x^2-\dfrac{1}{2}x^3$
