Answer
$1-x-\dfrac{1}{2}x^2-\dfrac{1}{2}x^3+.....$
Work Step by Step
Formula to find the binomial series is:
$(1+x)^m=1+\Sigma_{k=1}^\infty \dbinom{m}{k}x^k$
Here, $\dbinom{m}{k}=\dfrac{m(m-1)(m-2).....(m-k+1)}{k!}$
Now, $(1-2x)^{1/2}=1+(\dfrac{1}{2})(-2x)+\dfrac{(\dfrac{1}{2})(-\dfrac{1}{2})(-2x)^2}{2!}+\dfrac{(\dfrac{1}{2})(-\dfrac{1}{2})(-\dfrac{3}{2})(-2x)^3}{3!}+...$
Thus, the first four terms are: $1-x-\dfrac{1}{2}x^2-\dfrac{1}{2}x^3+.....$
