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