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.

The key here is to look at the constant(!) differences. In $1+2+3+4+...n$ it is $d=1$, and the last number is, calling the first number $a$,  $a+(n-1)*d=1+n-1=n$. In your next example the difference is $d=11$, and with $a=11$ you have the last number $a+(n-1)*d=11+(n-1)*11 = 11n $ and the sum can then analogically be computed like the first sum.
