Binomial summation problem 
Show that 
  $$ \frac{\binom{n}{0}}{1} - \frac{\binom{n}{1}}{4} +\dots + (-1)^n \frac{\binom{n}{n}}{3n+1} = \frac{3^n \cdot n!}{ 1\cdot 4\cdot 7\cdots(3n+1)}.$$

I don't know how to proceed in such type of problems. Any help or hint will be much appreciated.

Note $$(1-x^{3})^{n} = \binom{n}{0} x^{0} - \binom{n}{1}x^{3} + \binom{n}{2}x^{6} + \cdots +(-1)^{n}\binom{n}{n}x^{3n}$$ Then what you need is $$\int_{0}^{1}(1-x^{3})^{n} \ dx = \frac{\Gamma\frac{4}{3} \cdot n!}{\Gamma\left(n+\frac{4}{3}\right)}$$
