Evaluating $\sum_{r=1}^{3n-1}\dfrac{(-1)^{r-1}\cdot r}{\binom{3n}r}$ 
$$\sum_{r=1}^{3n-1}\dfrac{(-1)^{r-1}\cdot r}{\binom{3n}r}, n \in 2k, k\in \mathbb{Z^+}$$

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...
 A: 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 : $$ \sum_{r=0}^{3n-1}{\frac{\left(-1\right)^{r-1}r}{\binom{3n}{r}}}=\frac{3n}{3n+2}$$
A: With $n=2k$ we have:
$$\begin{align}
\sum_{r=1}^{6k-1}\dfrac{(-1)^{r-1}\cdot r}{\binom{6k}r}
&=\sum_{r=1}^{3k-1}\dfrac{(-1)^{r-1}\cdot r}{\binom{6k}r}
+(-1)^{3k-1}\frac{3k}{\binom{6k}{3k}}
+\sum_{r=3k}^{6k-1}\dfrac{(-1)^{r-1}\cdot r}{\binom{6k}r}\\
&=(-1)^{3k-1}\frac{3k}{\binom{6k}{3k}}
+\sum_{r=1}^{3k-1}\left[\dfrac{(-1)^{r-1}\cdot r}{\binom{6k}r}
+\dfrac{(-1)^{r-1}\cdot (6k-r)}{\binom{6k}r}\right]\\
&=(-1)^{3k-1}\frac{3k}{\binom{6k}{3k}}+6k\sum_{r=1}^{3k-1}\dfrac{(-1)^{r-1}}{\binom{6k}r}\tag1\\
&=(-1)^{3k-1}\frac{3k}{\binom{6k}{3k}}+6k\frac{6k+1}{6k+2}\left[\frac1{\binom{6k+1}1}+\dfrac{(-1)^{3k-2}}{\binom{6k+1}{3k}}\right]\\
&=\frac{3k}{3k+1}.
\end{align}$$
where we used the fact that the sum in (1) is telescopic due to identity:
$$
\dfrac1{\binom{m}r}=\frac{m+1}{m+2}\left[\frac1{\binom{m+1}r}+\frac1{\binom{m+1}{r+1}}\right].
$$
A: This is true more generally if we replace $3n$ by $n$.
Wolfram|Alpha returns this closed form for the partial sums:
$$
\sum_{r=1}^{k-1}\frac{(-1)^{r-1}r}{\binom nr}=\frac{(n+1)\binom nk-(-1)^k\left(k^2(n+2)-k(n^2+3n+3)+n+1\right)}{(n+2)(n+3)\binom nk}\;.
$$
This should be provable with induction over $k$. Substituting $k=n$ yields
$$
\sum_{r=1}^{n-1}\frac{(-1)^{r-1}r}{\binom nr}=\frac{(n+1)+(-1)^n(n^2+2n-1)}{(n+2)(n+3)}\;.
$$
For even $n$, this simplifies to $\frac n{n+2}$.
