How to show the double sum identity $\sum ^{n}_{i=1}\sum ^{i}_{j=1}i-j$ = $\dfrac {1}{6}n\left( n-1\right) \left( n+1\right) $ I am not sure how to go about showing this: $\sum ^{n}_{i=1}\sum ^{i}_{j=1}i-j$ = $\dfrac {1}{6}n\left( n-1\right) \left( n+1\right) $
It is a bit like the formula for $\sum ^{n}_{i=1}i^{2}$
and this $\sum_{i=1}^n \sum_{j=1}^i \frac{i-j}{nm} + \sum_{i=1}^{n} \sum_{j=i}^m \frac{j-i}{nm} = \frac{2 n^2 - 3 n m + 3 m^2  - 2}{6m}$
As I really do not know how to proceed.

I would suggest an inductive proof to that identity. Assume the induction hipotesis
$$
\sum^{n}_{i=1}\sum^{i}_{j=1}i-j= \dfrac{1}{6}n\left( n-1\right) \left( n+1\right).
$$
It's easy to verify identity for $n=1$, $n=2$ and $n=3$. Consider the scheme.
$$
\begin{array}{rl}
\sum^{n}_{i=1}\sum^{i}_{j=1}i-j=&0
\\
+&0+1 
\\
+&0+1+2
\\
+&0+1+2+3
\\
+&0+1+2+3+4
\\
&\vdots \;\;\;\,\vdots \;\;\;\,\vdots \;\;\;\, \vdots \;\;\;\, \vdots \;\;\;\, \ddots
\\
+&0+1+2+3+4+\cdots +i
\\
&\vdots \;\;\;\,\vdots \;\;\;\,\vdots \;\;\;\, \vdots \;\;\;\,\vdots \;\;\;\, \quad \;\;\;\, \vdots \;\;\;\,\ddots
\\
+&0+1+2+3+4+\cdots +i+\cdots+(n-1)
\end{array}
$$
Note by scheme that 
\begin{align}
\sum^{n+1}_{i=1}\sum^{i}_{j=1}i-j &= \left[\sum^{n}_{i=1}\sum^{i}_{j=1}i-j\right]+\big[ 1+2+3+\ldots +n\big]
\end{align}
By induction hipotesis we have 
\begin{align}
\sum^{n+1}_{i=1}\sum^{i}_{j=1}i-j &= \left[\dfrac{1}{6}\left( n-1\right)n\left( n+1\right)\right]+\big[ 1+2+3+\ldots +n\big]
\end{align}
And since we know that $\dfrac{1}{2}n(n+1)$ is the result of the sum $1+2+3+\ldots +n$ we have
\begin{align}
\sum^{n+1}_{i=1}\sum^{i}_{j=1}i-j &= \dfrac{1}{6}( n-1)n(n+1)+ \dfrac{1}{2}n(n+1)
\\
&= \dfrac{1}{6}(n-1)n(n+1)+ \dfrac{3}{6}n(n+1)
\\
&= \Big[(n-1)+ 3\Big]\dfrac{1}{6}n( n+1)
\\
&= \dfrac{1}{6}n\cdot (n+1)(n+2)
\\
&= \dfrac{1}{6}[\color{red}{(n+1)}-1]\cdot [\color{red}{(n+1)}][\color{red}{(n+1)}+1]
\end{align}
