Answer
$1-\dfrac{3x}{2}+\dfrac{15 x^2}{8}-\dfrac{35x^3}{16}+...$
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 $\dfrac{-3}{2}$ for $x$ in the above form through the first four terms.
Therefore, we have:
$(1+x)^{-3/2}=1+\dfrac{-3x}{2}+\dfrac{\dfrac{-3}{2}(\dfrac{-3}{2}-1)x^2}{2!}x^2+\dfrac{\dfrac{-3}{2}(\dfrac{-3}{2}-1)(\dfrac{-3}{2}-2)}{3!}x^3+...\\=1-\dfrac{3x}{2}+\dfrac{15 x^2}{8}-\dfrac{35x^3}{16}+...$
