What is the formula for the binomial expansion of 1/sqrt(1-x) in series form?

[Greger]

how do you find the binomial expansion of 1/sqrt(1-x) in series form?

i know what the term by term expansion is but I'm trying to find the series representation,

the closest i have found involved double factorials and I'm sure there's an easier representation,

i've been trying to use the binomial theorem but i get fractional factorials which just give ∞.

is there some formula that i haven't been able to find to apply to this?

 

[Bohrok]

Assuming first of all that you have the correct coefficients for the binomial expansion of 1/√(1+x) as below,
$$\underbrace{\frac{1}{2}}_{n=1}, \underbrace{\frac{1\cdot 3}{2\cdot 4}}_{n=2}, \underbrace{\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}_{n=3}, \underbrace{\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}}_{n=4}, ...$$there is a way to get a nice closed formula for the coefficients without double factorials

If you look at the denominator of the fourth term for example,
2·4·6·8 = 2(1)·2(2)·2(3)·2(4) = 2⁴·4!
then the general form for the denominator is 2ⁿn!Now for the numerator of the fourth term,
1·3·5·7 = (1·2·3·4·5·6·7·8)/(2·4·6·8) = 8!/(2⁴·4!)
and so the general form for the numerator is (2n)!/(2ⁿn!)

Combining them together gives you
$$\frac{(2n)!}{(2^nn!)^2}$$