Do you have any other way of deriving the Taylor-Maclaurin series of $\sqrt{1-x}$? Here is mine!
PROOF
The Binomial Series Theorem is given by
\begin{align} (1+x)^\alpha=\sum^{\infty}_{k=0}{\alpha\choose k }x^k\;\;\text{for}\;\alpha \in \Bbb{R}\;\text{and}\;|x|<1.\end{align}
For $\alpha=\frac12$, we have from the Binomial series, \begin{align} (1-x)^{\frac12}=\sum^{\infty}_{k=0}(-1)^{k}{\frac12\choose k }x^{k}.\end{align} For fixed $k\in \Bbb{N},$ \begin{align} {\frac12\choose k }&=\frac{\frac{1}{2 }\left(\frac{1}{2 }-1\right)\cdots \left(\frac{1}{2 }-k+1\right)}{k !}\\&=\frac{(-1)^{k-1}}{2^k}\frac{1\cdot 3\cdot 5\cdots (2k-3)}{k !}\\&=\frac{(-1)^{k-1}}{2^k}\frac{(2k-2)!}{ 2^{k-1}(k-1)!k !}\\&=\frac{(-1)^{k-1}}{2^{2k-1} }\cdot\frac{1}{ k(k-1)!}\cdot\frac{(2k-2)!}{ (k-1)!} \\&=\frac{(-1)^{k-1}}{2^{2k-1}k}\frac{(2k-2)!}{ (k-1) ! \left[(2k-2)-(k-1)\right]!}\\&=\frac{(-1)^{k-1}}{2^{2k-1}k}{2k-2\choose k-1}.\end{align} Thus,
\begin{align} (1-x)^{\frac12}&=1+\sum^{\infty}_{k=1}(-1)^{k}{\frac12\choose k }x^{k},\;\;\text{where}\;(-1)^0\cdot^\frac12 C_{0}=1,\\&=1+\sum^{\infty}_{k=1}\frac{(-1)^{2k-1}}{2^{2k-1}k}{2k-2\choose k-1}x^{k}\\&=1-\sum^{\infty}_{k=1}\frac{2}{4^{k}k}{2k-2\choose k-1}x^{k}\\&=1-\sum^{\infty}_{k=1}\frac{2}{k}{2k-2\choose k-1}\left(\frac{x}{4}\right)^{k}\\&=1-\sum^{\infty}_{k=0}\frac{2}{k+1}{2k\choose k}\left(\frac{x}{4}\right)^{k+1}.\end{align}