Find the coefficient of $x^{4}$ in $(1+x)^{\frac{1}{3}}$ Find the coefficient of $x^{4}$ in $(1+x)^{\frac{1}{3}}$ 
I try to use the general binomial coefficient to do this
so I will first find ${{\frac{1}{3}} \choose k}$ 
which is equal to ${\frac{(\frac{1}{3})(\frac{1}{3}-1) ... ({\frac{1}{3}-k+1})}{k!}}$
and I take out $(-1)^k$ and I will get ${\frac{(\frac{-1}{3})(\frac{-1}{3}+1) ... ({\frac{1}{-3}+k-1})}{k!}}$ which is equal $(-1)^k {{\frac{1}{-3}+k-1} \choose k!}$
back to $x^{4}$ in $(1+x)^{\frac{1}{3}}$ 
$(1+x)^{\frac{1}{3}} = {\sum_k{{\frac{1}{-3}+k-1}\choose{k!}}(-1)^kx^k}$
So when to find $[x^4]$ it is just equal to  ${{\frac{1}{-3}+4-1} \choose {4!} } (-1)^4$ ?

We have $(1+x)^\alpha = \sum_{k=0}^\infty \binom{\frac{1}{3}}{k} x^k$, for $|x|<1$.
Hence the coefficient of $x^4$ is $\binom{\frac{1}{3}}{4} =\frac{(\frac{1}{3}) (\frac{-2}{3}) (\frac{-5}{3}) (\frac{-8}{3}) }{4!} = - \frac{10}{243}$.
