Answer
$1+\dfrac{1}{2}x-\dfrac{1}{8}x^2+\dfrac{1}{16}x^3+...$
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, $(1+x)^{1/2}=1+(\dfrac{1}{2})x+\dfrac{(\dfrac{1}{2})(-\dfrac{1}{2})x^2}{2!}+\dfrac{(\dfrac{1}{2})(-\dfrac{1}{2})(-\dfrac{3}{2})x^3}{3!}+\dfrac{(\dfrac{1}{2})(-\dfrac{1}{2})(-\dfrac{3}{2})(-\dfrac{5}{2})x^4}{4!}+...=1+\dfrac{1}{2}x-\dfrac{1}{8}x^2+\dfrac{1}{16}x^3+...$
Hence, we have the first four terms are: $1+\dfrac{1}{2}x-\dfrac{1}{8}x^2+\dfrac{1}{16}x^3+...$
