$i+j-1$ tokens on row $i$ column $j$ of square grid Consider an $n\times n$ grid, with rows and columns numbered from $1$ to $n$. On the cell in row $i$ column $j$, I place $i+j-1$ tokens. How many tokens are on the grid in total?
I have done the double sum directly:
$$\sum_{i=1}^n\sum_{j=1}^n(i+j-1)=\sum_{i=1}^n\left[n(i-1)+\frac{n(n+1)}{2}\right]=\frac{n^2(n-1)}{2}+\frac{n^2(n+1)}{2}=n^3.$$
With an answer as neat as this, I'm wondering if there's a simple (perhaps geometric) interpretation/argument?
 A: Picture for $n=6$:

Take a horizontal cut at the height of $n$ tokens. The tokens above the cut are all in the bottom right corner. Flip this shape over and it will fit snugly into the top left corner to form an $n\times n\times n$ cube of tokens.
A: Perhaps the following is an overkill, but explains how to backtrack arguments with combinatorials. Call $[n]=\{1,2,\cdots ,n\}.$ You can backtrack the argument. Suppose you want to count the number of functions from $[3]$ to $[n].$ this is clearly $n^3$ now, fix the image of $1$ so
$$n^3=n\cdot n^2=n\cdot \left (2\binom{n}{2}+\binom{n}{1}\right ),$$
where the second equality is given by either you choose two images or one for $2,3.$
Now, one copy of $\binom{n}{2}$ you combined it with $\binom{n}{1}$ getting $n^3=n\left (\binom{n}{2}+\binom{n+1}{2}\right)$ using the recursion of pascal interpretation(basically if $n+1$ belongs to the subset is because you were fixing just one image. The function is constant in $2,3$). Now you use the hockey stick interpretation(this is getting the biggest element in the subset and then the other ones.) to get
$$n^3=\sum _{j=1}^{n-1}\left (n\cdot j+\binom{n+1}{2}\right )=\sum _{j=1}^{n}\left (n\cdot (j-1)+\binom{n+1}{2}\right ),$$
again Hockeystick to get
$$n^3=\sum _{j=1}^{n}\sum _{i=1}^nj-1+i,$$
so the whole interpretation becomes as follows:
Suppose you have a function $f:[3]\longrightarrow [n]$ then either $f(2)>f(3)$ or $f(2)\leq f(3).$ If it is the first call $i$ the image of $1$ and call $j$ the image of $f(2)$ then you have  $j-1$ options for $f(3)$ giving you $\sum _{i,j}j-1.$ If it is the other case, call $j$ the image $f(1)$ and call $i+1$ the image of $3,$ then you have $i$ options for $f(2).$ If $i+1=n+1,$ then your function will be constant on $2,3$ and you have $n$ options to pick this number giving you $\sum _{i,j}i.$ Because this options are disjoint, you add them.
