Asymptotic of a sum involving binomial coefficients Could you help me to find an asymptotic for this sum?
$$ \sum_{k=0}^{n - 1} (-1)^k {n \choose k} {3n - k - 1 \choose 2n - k}  = {n \choose 0} {3n - 1 \choose 2n} - {n \choose 1} {3n - 2 \choose 2n - 1}  + ... + (-1)^{n-1} {n \choose n-1} {2n \choose n + 1}  $$
I have tried to write binomials through factorials and work with it, but it seems like not right way to evaluate an asymptotic.
Thank you for your help:)
 A: $$\begin{align}
\sum_{k=0}^{n-1}(-1)^k\binom nk \color{blue}{\binom {3n-k-1}{2n-k}}
&=\sum_{k=0}^{n-1}(-1)^k\binom nk\color{blue}{(-1)^{2n-k}\binom {-n}{2n-k}}
\qquad&\color{blue}{\text{(upper negation)}}\\
&=\color{green}{\left[\sum_{k=0}^{n}\binom nk\binom {-n}{2n-k}\right]}-\color{orange}{\binom nk\binom {-n}{2n-k}\Biggr|_{k=n}}\\
&=\color{green}{\binom 0{2n}}-\color{orange}{\binom nn\binom {-n}{n}}
\qquad&\color{green}{\text{(Vandermonde)}}\\
&=-\color{orange}{(-1)^n\binom{2n-1}n}
\qquad&\color{orange}{\text{(upper negation)}}\\
&=(-1)^{n-1}\binom{2n-1}n=(-1)^{n-1}\binom{2n-1}{n-1}\quad\blacksquare
\end{align}$$
A: Suppose we first seek to evaluate the somewhat more general
$$S_m(n) = \sum_{k=0}^n {n\choose k} (-1)^k
{mn-k-1\choose 2n-k}$$
with $m$ an integer parameter.
Now introduce
$${mn-k-1\choose 2n-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-k+1}} (1+z)^{mn-k-1} \; dz.$$

We thus get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{mn-1}}{z^{2n+1}}
\sum_{k=0}^n {n\choose k} (-1)^k \frac{z^k}{(1+z)^k}  \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{mn-1}}{z^{2n+1}}
\left(1-\frac{z}{1+z}\right)^n\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{mn-n-1}}{z^{2n+1}} \; dz
\\ = {mn-n-1\choose 2n}.$$
We see that this is zero when $m=2$ and $m=3$ and non-zero otherwise.
In the problem being asked we are computing
(the last term is not being included)
$$S_3(n) - (-1)^n {2n-1\choose n}$$ 
so we obtain
$$(-1)^{n+1} {2n-1\choose n}.$$
This  can be  treated with  the  asymptotic for  the central  binomial
coefficient. We get
$$(-1)^{n+1} \frac{n}{2n} {2n\choose n}
= \frac{1}{2} (-1)^{n+1} {2n\choose n}.$$
The central binomial coefficient is 
OEIS A000984
and is asymptotic to
$$\frac{4^n}{\sqrt{\pi n}}.$$
A: Hint: Use $\displaystyle{a-1\choose b}=(-1)^b~{b-a\choose b}$ in conjunction with Vandermonde's identity. Also, notice 
that the upper summation limit is $n-1$ instead of n.
