<s>[INST] Taylor series for $\sqrt{1+x^2}$ I want to expand 
$$f(x)=\sqrt{1+x^2}$$ in powers of $x-2$ 
I started by getting the maclaurin series
$$\sqrt{1+x}=1+\frac{1}{2}x+\frac{1}{2} \left( \frac{-1}{2} \right) \frac{x^2}{2!} + \frac{1}{2} \left( \frac{-1}{2}\right) \left(\frac{-3}{2}\right)\frac{x^3}{3!}$$
$$\sqrt{1+x^2}=1+\frac{1}{2}x^2+\frac{1}{2}\left(\frac{-1}{2}\right)\frac{x^4}{2!}+\frac{1}{2} \left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right)\frac{x^6}{3!}$$
Then
$$\sqrt{1+x^2}=\sqrt{1+(x-2)^2-4+4x}=\sqrt{(x-2)^2-3+4x}=\sqrt{(x-2)^2-3+4(x-2)+8}$$
I could not complete , what can we do then ?
(I know that we can differentiate the function and substitute in the Taylor formula , but I want a shorter way)
for example : 
if we want to expand $$g(x)=\frac{1}{1-x}$$ around $x=2$ we can start by
\begin{align}
g(x) & =\frac{1}{1-x}=\frac{1}{1-(x-2)-2}=\frac{-1}{1+(x+2)} \\[10pt]
& =-[1-(x-2)+(x-2)^2-(x-2)^3+\cdots]
\end{align}
So  I want to get convert $f(x)$ to a form that we can write its expansion without getting derivatives, like I did with $g(x)$ above
 [/INST] The Generalized binomial formula for $a\in \Bbb R\setminus \{\Bbb N\}$ says 
$$(1+x)^a= \sum_{k=1}^{\infty}{a\choose k} x^k$$
where: $$ \color{blue}{{a\choose k} =\frac{a(a-1)\cdots(a-k+1)}{k!}}$$
see here https://en.wikipedia.org/wiki/Binomial_series
just take $a=1/2$ and replace $x$ by $x^2$. 
</s>