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.
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Let $$a_n:=\frac{\binom{n}{0}}{1} - \frac{\binom{n}{1}}{4} +... + (-1)^n \frac{\binom{n}{n}}{3n+1} =\sum_{k=0}^n\frac{(-1)^k\binom{n}{k}}{3k+1}$$ Then for $n\geq 1$, \begin{align} a_n&=1+\sum_{k=1}^n\frac{(-1)^k\frac{n}{k}\binom{n-1}{k-1}}{3k+1} =1+n\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}\left(\frac{1}{k}-\frac{3}{3k+1}\right) \\ &= 1+n\sum_{k=1}^n\frac{(-1)^k\binom{n-1}{k-1}}{k}-3n\sum_{k=1}^n\frac{(-1)^k\binom{n-1}{k-1}}{3k+1}\\ &=\sum_{k=0}^n(-1)^k\binom{n}{k}-3n\sum_{k=1}^n\frac{(-1)^k\left(\binom{n}{k}-\binom{n-1}{k}\right)}{3k+1}\\ &=(1-1)^n-3n(a_n-1)+3n(a_{n-1}-1)=-3na_n+3na_{n-1} \end{align} which implies that $$a_n=\frac{3n a_{n-1}}{3n+1}$$ and the desired identity easily follows $$a_n=\frac{3n a_{n-1}}{3n+1}=\frac{3^2n(n-1) a_{n-2}}{(3n+1)(3(n-1)+1)}=\dots=\frac{3^nn!}{1\cdot 4\cdot 7\cdots(3n+1)}.$$
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 $$\int_{0}^{1}(1-x^{3})^{n} \ dx = \frac{\Gamma\frac{4}{3} \cdot n!}{\Gamma\left(n+\frac{4}{3}\right)}$$