Answer
$1+\dfrac{x}{4}-\dfrac{3}{32}x^2+\dfrac{7}{128}x^3+...$
Work Step by Step
We have the Maclaurin series for $(1+x)^a$ as follows
$(1+x)^a=1+ax+\dfrac{a(a-1)x^2}{2!}x^2+\dfrac{a(a-1)(a-2)}{3!}x^3+...$
Replace $\dfrac{1}{4}$ for $x$ in the above form through four terms.
Therefore, we have:
$(1+x)^{1/4}=1+\dfrac{x}{4}+\dfrac{\dfrac{1}{4}(\dfrac{1}{4}-1)x^2}{2!}x^2+\dfrac{\dfrac{1}{4}(\dfrac{1}{4}-1)(\dfrac{1}{4}-2)}{3!}x^3+...\\=1+\dfrac{x}{4}-\dfrac{3}{32}x^2+\dfrac{7}{128}x^3+...$
