Expansion of $(a+b)^{1/3}$ with binomial theorem

Question

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 :)

Answer 1

Binomial theorem works for non negative integer $n$.

What we can do is use the binomial series expansion for

$$(a+b)^{\frac13}=a^\frac13(1+b/a)^\frac13=a^\frac13\left(1+\frac{b}{3a}-\frac{b^2}{9a^2}+...\right)$$

Answer 2

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?