$$\sum_{r = 1}^{3n-1}\left(-1\right)^{r – 1}\,\,\dfrac{r}{{3n \choose r}},\quad n \in 2k,\ k\in \mathbb{Z^+}$$
Answer given (much simpler than expected)
$\dfrac{3n}{3n+2}$
I tried adding and subtracting 1 to $r$ so could use $\dfrac{\binom{n}r}{r+1}=\dfrac{\binom{n+1}{r+1}}{n+1}$, but didn't prove to be useful. I know summation and double summation of binomial coefficients to quite to good extent. If you could help…
Best Answer
Let's start by the following : \begin{aligned}\frac{1}{\binom{3n}{r}}=\left(3n+1\right)\int_{0}^{1}{x^{r}\left(1-x\right)^{3n-r}\,\mathrm{d}x}\end{aligned}
Since $\left(\forall x\in\left[0,1\right]\right) $, we have : $$ \sum_{r=0}^{3n-1}{r\left(-1\right)^{r-1}x^{r}\left(1-x\right)^{3n-r}}=\left(-1\right)^{n+1}x^{3n+1}\left(1-x\right)+3\left(-1\right)^{n}n x^{3n}\left(1-x\right)+x\left(1-x\right)^{3n+1} $$
It can be proven by differentiating the formula giving the sum of consecutive terms of a geometric sequence.
Thus, \begin{aligned}\scriptsize \sum_{r=0}^{3n-1}{\frac{\left(-1\right)^{r-1}r}{\binom{3n}{r}}}&\scriptsize=\left(3n+1\right)\int_{0}^{1}{\sum_{r=0}^{3n-1}{r\left(-1\right)^{r-1}x^{r}\left(1-x\right)^{3n-r}}\,\mathrm{d}x}\\ &\scriptsize=\left(-1\right)^{n+1}\left(3n+1\right)\int_{0}^{1}{x^{3n+1}\left(1-x\right)\mathrm{d}x}+3\left(-1\right)^{n}n\left(3n+1\right)\int_{0}^{1}{x^{3n}\left(1-x\right)\mathrm{d}x}+\left(3n+1\right)\int_{0}^{1}{x\left(1-x\right)^{3n+1}\,\mathrm{d}x}\\ &\scriptsize=\frac{\left(-1\right)^{n+1}\left(3n+1\right)}{\left(3n+2\right)\left(3n+3\right)}+\frac{3\left(-1\right)^{n}n}{3n+2}+\frac{\left(3n+1\right)}{\left(3n+2\right)\left(3n+3\right)} \end{aligned} And hence, if your $ n $ is even, we can get rid of the $ \left(-1\right)^{n} $ and get some cancellations to end with : $$ \sum_{r=0}^{3n-1}{\frac{\left(-1\right)^{r-1}r}{\binom{3n}{r}}}=\frac{3n}{3n+2}$$