The summation to be evaluated is $$\sum_{k=0}^n (-1)^{k-1} \frac{k}{^{3n}C_k} \cdotp$$
I expanded the denominator (using ${^{3n}C_k}=\frac{3n!}{k!({3n-k})!}$) and took $3n!$ outside the summation and I obtained this expression
$$\frac{1}{3n!} \sum_{k=0}^{3n-1} (-1)^{k-1}k!(3n-k)!k.$$
I thought this summation can be solved using the method of telescoping, but I could not do so.
What is the actual procedure to evaluate this summation?
Update: I used wolfram alpha to compute the above summation and saw that the solution is complicated $$ \sum_{k=0}^n \frac{(-1)^{k-1} k}{\binom{3n}{k}} = \frac{(3n+1) \binom{3n}{n+1} - 2 (-1)^n n (3 n^2 + 5n + 1)} {3 (n+1) (3n+2) \binom{3n}{n+1}} \cdotp $$
However, even with the solution being at hand, I am no closer to solving this misleadingly simple looking sum on pen and paper yet.