Consecutive integers sum with different steps First of all: beginner here, sorry if this is trivial.
We know that $ 1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2 $ .
My question is: what if instead of moving by 1, we moved by an arbitrary number, say 3 or 11? $ 11+22+33+44+\ldots+11n = $ ?
The way I've understood the usual formula is that the first number plus the last equals the second number plus second to last, and so on.
In this case, this is also true but I can't seem to find a way to generalize it.

There's a simple rule for linear progressions: The sum equals the number of items, times the average of the first and last item. So in this case n (11n + 11) / 2. 
If you added 7, 17, 27, 37 and so on, the first item would be 7, the last would be 10n - 3, the average (10n - 3 + 7) / 2 = 5n + 2, so the sum would be n (5n + 2). 
