How to expand $\sqrt{x^6+1}$ using Maclaurin's series The expansion would be
$\sum_{n=0}^\infty$$\frac{1}{2}\choose n $$x^{6n}$
How to evaluate binomial coefficient with rational numbers?
If $\frac{1}{2}\choose n $=$2n\choose n $$\times \frac{(-1)^{n+1}}{2^{2n}(2n-1)}$ what would be the expression for binomial coefficient if rational number is $\frac{3}{2}$ instead of $\frac{1}{2}$?
This is one example of Macclaurin expansion for the above expression:
$$1+\frac{1}{2x^6}+r(\frac{1}{x^{11}})$$
Can someone show the steps for expanding the expression, and how to generate remain?
 A: The generalised binomial coefficient $\,\dbinom{\tfrac12}n$ is simply:
$$\frac{\cfrac12\Bigl(\cfrac12-1\Bigr)\Bigl(\cfrac12-2\Bigr)\dots\Bigl(\cfrac12-n+1\Bigr)}{n\,!}.$$
A: If you want an alternative representation, one can simply recall that 
$$\frac{d^n(1+x)^{1/2}}{dx^n}=\frac{(-1)^{n+1}(2n-3)!!}{2^n}(1+x)^{-(2n-1)/2}$$
for $n\ge 2$, where $(2n-3)!!=1\cdot3\cdot5\cdots (2n-3)$ is the double factorical.  Then, we have
$$(1+x^6)^{1/2}=1+\frac{x^6}{2}+\sum_{n=2}^{\infty}\frac{(-1)^{n+1}(2n-3)!!}{2^nn!}x^{6n}\tag 1$$
Note that we can write the double factorial in terms of single factorials as
$$\begin{align}
(2n-3)!!&=\frac{(2n)!}{(2n)(2n-1)(2n-2)(2n-4)\cdots 2}\\\\
&=\frac{(2n)!}{(2n-1)2^nn!} \tag 2
\end{align}$$
Substituting $(2)$ into $(1)$ gives the expansion 
$$\begin{align}
(1+x^6)^{1/2}&=1+\frac{x^6}{2}+\sum_{n=2}^{\infty}\frac{(-1)^{n+1}(2n)!}{4^n(2n-1)(n!)^2}x^{6n}\\\\
&=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}(2n)!}{4^n(2n-1)(n!)^2}x^{6n}
\end{align}$$

NOTE:
Another way of writing the generalized binomial coefficient for $\binom{1/2}{n}$ is thus
$$\binom{1/2}{n}=\frac{(-1)^{n+1}(2n)!}{4^n(2n-1)(n!)^2}$$
A: When $n$ and $m$ are positive integers, the binomial coefficient $\binom{n}{m}$ is defined as:
$$ \binom{n}{m}=\frac{n!}{m!(n-m)!}\tag{1}$$
hence by replacing $a!$ with $\Gamma(a+1)$ it is natural to consider the extended definition:
$$\binom{n}{m}=\frac{\Gamma(n+1)}{\Gamma(m+1)\,\Gamma(n-m+1)}\tag{2}$$
from which it follows that:
$$\binom{\frac{1}{2}}{n}=\frac{\Gamma(3/2)}{n!\Gamma(3/2-n)}=\frac{\frac{1}{2}\cdot\left(-\frac{1}{2}\right)\cdot\ldots\cdot\left(\frac{3}{2}-n\right)}{n!}=\frac{1}{4^n}\binom{2n}{n}\frac{(-1)^{n+1}}{2n-1}\tag{3}$$
as you stated. Since:
$$ \frac{\binom{n+1}{m}}{\binom{n}{m}}=\frac{n+1}{n+1-m}\tag{4}$$
it follows that:
$$ \binom{\frac{3}{2}}{n}=\frac{\frac{3}{2}}{\frac{3}{2}-n}\binom{\frac{1}{2}}{n}=\frac{3}{4^n}\binom{2n}{n}\frac{(-1)^n}{(2n-1)(2n-3)}.\tag{5}$$
