How can I simplify: $\sum^{n-1}_{i=1}\sum^{n}_{j=i+1}\sum^{j}_{k=1} 1?$ I simplified the most inner sum to: $j$. So, now I have:
$$\sum^{n-1}_{i=1}\sum^{n}_{j=i+1}j$$
I'm not sure if the following is correct but here is what I am thinking. I can re-index the inner sum by letting $m = j - i$. So, now I have
$$\sum^{n-1}_{i=1}\sum^{n}_{m=1} (j - m + 1)$$
Then the inner sum becomes:
$$\frac{(j-m+1)(j - m + 2)}{2}$$
So now I'm left with:
$$\sum^{n-1}_{i=1} \frac{(j-m+1)(j - m + 2)}{2} $$
Since, $j - m = i$, I have:
$$\sum^{n-1}_{i=1} \frac{(i+1)(i + 2)}{2} $$
Assuming the above is correct (for whatever reason, it doesn't feel right) then the only issue I am having is the pesky $n-1$ at the top. Can I do a change of variable on that as well and if I do then what would I have to subtract (or add) later? If that's not a good method and what's a good way to go about figuring this out? 
Thanks a lot! 
 A: I am afraid that you have made a mistake somewhere in your work as the final sum does not evaluate to the same value as the initial sum.
Note: $\sum_{i=1}^n i=\frac{n(n+1)}{2}.\quad\!\!$ Thus, $\quad\!\!\!\sum_{i=a}^ni=\frac{(n+a)(n-a+1)}{2}=\sum_{i=1}^ni -\sum_{i=1}^{a-1}=\frac{n(n+1)}{2}-\frac{a(a-1)}{2}$ 
An easier way to approach this might be to evaluate the inner sum: 
$$\sum_{j=i+1}^nj=\dfrac{(n+i+1)(n-i)}{2}$$
Then, since $(n+i+1)(n-i)=n^2+n-i-i^2$, we have 
$$\sum_{i=1}^{n-1} \dfrac{n^2+n-i-i^2}{2}=\dfrac{1}{2}\left(n^2(n-1)+n(n-1)-\sum_{i=1}^{n-1}i-\sum_{i=1}^{n-1}i^2\right)$$

$$=\dfrac{1}{2}\left(n^2(n-1)+n(n-1)-\dfrac{1}{2}n(n-1)-\dfrac{1}{6}n(n-1)(2n-1)\right)=\dfrac{1}{3}n(n^2-1)$$

A: $$\begin{align}
\sum^{n-1}_{i=1}\sum^{n}_{j=i+1}\sum^{j}_{k=1} 1&=\sum_{j=2}^n\sum_{i=1}^{j-1}\sum_{k=1}^j1\\
&=\sum_{j=2}^nj(j-1)\\
&=2\sum_{j=2}^n \binom j2\\
&=2\binom {n+1}3\\
&=\frac 13 n(n^2-1)\qquad\blacksquare\end{align}$$
A: Just going to work this out with the OP's logic, making the necessary two change of variable corrections at the stumbling point (c.f. comment).
$\sum^{n-1}_{i=1}\sum^{n-i}_{m=1} (m + i) = \sum^{n-1}_{i=1}\left(\frac{(n-i)(n-i+1)}{2} + i(n-i)\right) = \sum^{n-1}_{i=1}(\frac{n^2+n-i-i^2}{2} )$
The accepted answer wraps thing up from here.
Although this might look like the accepted answer, it is more aligned with the OP's original intent; also, here we don't have to introduce a letter $a$.
