Represent the function in Maclaurin series and Taylor series around point $a$:
$$f(x) = \frac{1}{\sqrt{1 -x}}$$
I find the first few derivatives:
$$f = (1-x)^{-\frac{1}{2}}$$
$$f' = \frac{1.(1-x)^{-\frac{3}{2}}}{2}$$
$$f'' = \frac{1.3.(1-x)^{-\frac{5}{2}}}{2.2}$$
$$f''' = \frac{1.3.5.(1-x)^{-\frac{7}{2}}}{2.2.2}$$
so the pattern that I see is:
$$f^{(k)} = \frac{(2k - 1)!!(1-x)^{-\frac{(2k + 1)}{2}}}{2^{k}}$$
For Maclaurin series I just set $x$ to $0$ and get:
$$\sum_{n = 0}^{\infty}\frac{(2n - 1)!!x^n}{2^n n!}$$
However, for Taylor series expansion I'm not sure if I should substitute $x$ with $a$ in the derivatives and keep the $(1 - a)^{-\frac{(2k + 1)}{2}}$ term or do something different?
Hint: Taylor series is $$f(x)=f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3}+\cdots .$$
Maclaurin series is $$f(x)=f(0)+{\frac {f'(0)}{1!}}(x)+{\frac {f''(0)}{2!}}(x)^{2}+{\frac {f'''(0)}{3!}}(x)^{3}+\cdots .$$