For 1-dimension, it's simple.
$a_{n}=a_{1}+\sum_{i=2}^{n}(a_{i}-a_{i-1})$
But what would be the corresponding identity for 2-dimension?
In other words, if we put
$a_{n,m}=a_{1,1}+X$
then how can we express $X$ as an elegant sum of differences of $a_{ij}$ s?
The first thing that comes into mind is of course from (1,1) to (n,m), but I failed to put this into an elegant summation of differences of $a_{ij}$ s.