Getting a closed form from $\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1$ I need to get a closed form from
$$
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1
$$
Starting from the most outer summation, I got
$$
\sum_{k=1}^{j} 1 = j
$$
But now I don't know how to proceed with:
$$
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} j
$$
Could you guys please help me?
Thanks in advance.

Thanks @Winther for pointing out the previous mistake
\begin{equation}
 \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1
\end{equation}
We know that 
\begin{equation}
 \sum_{k=1}^{j} 1 = j
\end{equation}
So
\begin{equation}
 \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1 =
 \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} j
\end{equation}
\begin{equation}
 \sum_{j=i+1}^{n} j
 =
 \sum_{j=1}^{n} j
 -
 \sum_{j=1}^{i} j
 =
 \frac{n(n+1)}{2}
 -
 \frac{i(i+1)}{2}
\end{equation}
\begin{equation}
 \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1 =
 \sum_{i=1}^{n-1} ( \frac{n(n+1)}{2}
 -
 \frac{i(i+1)}{2})
 =
 \frac{n(n+1)(n-1)}{2}
 -
 \frac{1}{2}
 \sum_{i=1}^{n-1}
 i+i^2
\end{equation}
But
\begin{align}
 \sum_{i=1}^{n-1}
 i &= \frac{(n-1)n}{2} \\
 \sum_{i=1}^{n-1}
 i^2 &= \frac{(n-1)n(2n-1)}{6}
\end{align}
So
\begin{equation}
 \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1 =
 \frac{n(n+1)(n-1)}{2}
 -
 \frac{1}{2}
 (\frac{(n-1)n}{2} + \frac{(n-1)n(2n-1)}{6})
\end{equation}
Let's arrange
\begin{equation}
 \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1 =
\frac{6n(n+1)(n-1) - 3n(n-1) - n(n-1)(2n-1)}{12}
\end{equation}

We arrive at the most compact form,
  \begin{equation}
 \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1 =
\frac{(n-1)n(6n + 6 - 3 - 2n + 1)}{12}
=
\frac{(n-1)n(n + 1)}{3}
\end{equation}

