>That's how I got my answer, sort of.
>I graphed it and recognized that I was dealing with a polynomial. So I started by graphing y=x^2 and worked from there. I guess Edward's method was better than mine. I cheated using Excel. It's been years since high school.....

Here's a neat little proof I came up with last night (to beat a dying horse):
To keep this small, let n=5 for example sums shown, then we'll generalize. We want to find y = the solution, the sum of all the numbers. Then look at:
1
1+2
1+2+3
1+2+3+4
1+2+3+4+5
We want to sum the rows. Instead, let's sum the columns from right to left.
Then we have
y = n + 2(n-1) + 3(n-2) + 4(n-3) + ...generalizing from the rightmost column.

y = n + 2n + 3n + 4n + ...-2 - 6 - 12 - ...multiplying,rearranging
y = n(1+2+3+4+...) - 2(1+3+6+...) factoring. Now 1+3+6+... is almost y itself, but it's missing the last term, which is n(n+1)/2.
y = n*n(n+1)/2 - 2(1+3+6+...) we substitute using the known sum (1+2+3+...n)=n(n+1)/2. As noted, y is missing the nth term, so we need to add it and subtract it to the series 1+3+6+....The addition gives y:
y = n^2(n+1)/2 - 2(y - n(n+1)/2)
Now solve for y:
y = n^2(n+1)/2 -2y + n(n+1)
3y = n^2(n+1)/2 + n(n+1)
6y = n^2(n+1) + 2n(n+1) then factoring
y = n+1[n^2 + 2n]/6 = (n+1)n(n+2)/6 and QED, as they say.