How find this sum closed form? Have  this sum have close form  $$f(n)=\sum_{k=0}^{n-1}\left(\left(\sum_{i=0}^{k}(-1)^i\binom{n}{i}\right)\cdot\left(‌​\sum_{j=k+1}^{n}(-1)^j\binom{n}{j}\right)\right)$$
Maybe this sum can use integral to solve it? Thank you 
 A: This sum is actually quite tame if you take it step by step.
The sum comprising the first factor is:
$$\sum_{j=0}^{k}(-1)^j\binom{n}{j}=(-1)^k\binom{n-1}{k}.$$
And since $\sum_{j=0}^{n}(-1)^j\binom{n}{j}=0$, we immediately find the sum in the second factor to be:
$$\sum_{j=k+1}^{n}(-1)^j\binom{n}{j}=-(-1)^k\binom{n-1}{k}.$$
Hence, the sum for $f(n)$ can be written as a single finite binomial sum whose values are well known to be the (negatives of the) central binomial coefficients:
$$f(n)=-\sum_{k=0}^{n-1}\binom{n-1}{k}^2=-\binom{2(n-1)}{n-1}.$$
A: I've simply added a justification of the first identity in David H's answer.
Assume that $n\ge1$, so that $\binom{n-1}{n}=0$.
$$
\begin{align}
\sum_{j=0}^k(-1)^j\binom{n}{j}
&=\sum_{j=0}^k(-1)^j\left[\binom{n-1}{j}+\binom{n-1}{j-1}\right]\\
&=\sum_{j=0}^k(-1)^j\binom{n-1}{j}-\sum_{j=0}^{k-1}(-1)^j\binom{n-1}{j}\\
&=(-1)^k\binom{n-1}{k}\tag{1}
\end{align}
$$
Using $k=n$ in $(1)$ yields
$$
\begin{align}
\sum_{j=0}^n(-1)^j\binom{n}{j}
&=(-1)^n\binom{n-1}{n}\\
&=0\tag{2}
\end{align}
$$
Therefore, using Vandermonde's Inequality and $(1)$ and $(2)$ gives
$$
\begin{align}
&\sum_{k=0}^n\left(\sum_{i=0}^k(-1)^i\binom{n}{i}\right)\left(\sum_{j=k+1}^n(-1)^j\binom{n}{j}\right)\\
&=\sum_{k=0}^n(-1)^k\binom{n-1}{k}(-1)^{k+1}\binom{n-1}{k}\\
&=-\sum_{k=0}^n\binom{n-1}{k}\binom{n-1}{n-1-k}\\
&=-\binom{2n-2}{n-1}\tag{3}
\end{align}
$$
