Expansion of $(a+b)^{1/3}$ with binomial theorem I want to expand $(a+b)^{1/3}$ using the binomial theorem.
1) Is it possible to do that? If not, which expansion should I look up to do so?
2) If yes, how do I evaluate $\dbinom{n}{\frac{1}{3}}$?
Thanks :)
 A: Binomial theorem works for non negative integer $n$.
What we can do is use the binomial series expansion for
$$(a+b)^{\frac13}=a^\frac13(1+b/a)^\frac13=a^\frac13\left(1+\frac{b}{3a}-\frac{b^2}{9a^2}+...\right)$$
A: As User GIMUSI already told you, use his method to get a writing of that kind in order to use then:
$$(1 + x)^{\alpha}= \sum_{k = 0}^{+\infty} \binom{\alpha}{k} x^k$$
In your case $\alpha = 1/3$.
Notice that
$$\binom{1/3}{k} = \frac{\frac{1}{3}\left(\frac{1}{3} - 1\right)\ldots \left(\frac{1}{2} - k + 1\right)}{k!}$$
Thanks to the multiplicative rule.
For example:
$$\begin{align*}
\binom{1/2}n&=\frac{(1/2)^{\underline n}}{n!}\\
&=\frac{\left(\frac12\right)\left(-\frac12\right)\left(-\frac32\right)\dots\left(\frac{2n-3}2\right)}{n!}\\
&=(-1)^{n-1}\frac{(2n-3)!!}{2^nn!}\\
&=(-1)^{n-1}\frac{2^{n-1}(n-1)!(2n-3)!!}{2^{2n-1}n!(n-1)!}\\
&=(-1)^{n-1}\frac{(2n-2)!!(2n-3)!!}{2^{2n-1}n!(n-1)!}\\
&=\frac{(-1)^{n-1}}{2^{2n-1}n}\frac{(2n-2)!}{(n-1)!^2}\\
&=\frac{(-1)^{n-1}}{2^{2n-1}n}\binom{2n-2}{n-1}
\end{align*}$$
Can you procede in a similar way for your case? 
