As far as I know, there are two possible methods:
- Use that for $a \in \Bbb R${\left( {1 + x} \right)^a} = \sum\limits_{n = 0}^\infty {a \choose n}{x^n}
This will yield
$\sqrt{1 + 2x}= \sum\limits_{n = 0}^\infty {1/2 \choose n}2^n{x^n} $
Now we need to find a closed form of a fractional binomial coefficient. This can be done rather "empirically", and I hope you also arrive at
${1/2 \choose n}= {\left( { - 1} \right)^{n + 1}}\frac{{(2n-3)\cdots5\cdot3\cdot1}}{{2^n\cdot n!}} $
which gives you each coefficient of the expansion.
- Another procedure would be computing each derivative at $x=0$ and trying to find a pattern. If you do things right, you will arrive at
$f^{(n)}(0)= {\left( { - 1} \right)^{n + 1}}\left[ {\left( {2n - 3} \right) \cdots 5 \cdot 3 \cdot 1} \right]$
i.e. $\left\{ f^{(n)}(0) \right\}=\{1,1,-1,3,-3\cdot 5,3\cdot5\cdot7,\dots\}$