Evaluate $\frac{C_0}{1}-\frac{C_1}{4}+\frac{C_2}{7}-\cdots+\frac{(-1)^nC_n}{3n+1}$

Question

Evaluate

$$S=\frac{C_0}{1}-\frac{C_1}{4}+\frac{C_2}{7}-\cdots+\frac{(-1)^nC_n}{3n+1}$$ where

$C_r$ denotes $\binom{n}{r}$

Need a hint please

Answer 1

This looks like an integral: $$\frac1{m+1}=\int_0^1x^m\,dx$$ so $$S=\int_0^1\left(\sum_{r=0}^n(-1)^n{n\choose r}x^{3r}\right)\,dx$$ etc.

Answer 2

Putting together all the fragments:

$$ \begin{eqnarray*} \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{3k+1} &=& \int_{0}^{1}\sum_{k=0}^{n}\binom{n}{k}(-x^3)^k\,dx \\ &=& \int_{0}^{1}(1-x^3)^n\,dx \\ &\stackrel{x\mapsto z^{1/3}}{=}& \frac{1}{3}\int_{0}^{1}z^{-2/3}(1-z)^n\,dz\\&=&\frac{\Gamma(1/3)\,\Gamma(n+1)}{3\,\Gamma(n+4/3)}\\&=&\frac{n!}{\prod_{k=1}^{n}\left(k+\frac{1}{3}\right)}\\&=&\color{blue}{\frac{1}{\prod_{k=1}^{n}\left(1+\frac{1}{3k}\right)}}. \end{eqnarray*} $$

Answer 3

Let $P(x)=C_0-C_1x^3+\cdots+(-1)^nC_nx^{3n}=(1-x^3)^n.$

Now find $$S=\int_0^1P(x)dx.$$