n = 1:
For n = 1, clearly the sum is 1, right?
S = 1 * (1 + 2) / 2 = 1*2/2 = 1, CHECK
Now let's assume that for n, S is correct, then what is S for n + 1?
S(n) = n(n + 1) / 2
-->
S(n + 1) = S(n) + (n + 1)
If we can show that S(n + 1) = (n + 1)(n + 2) / 2, then we will have proven that if this formula is correct for n, then it is ALSO correct for (n + 1)...and since it is correct for 1, it is correct for 2...but since it's correct for 2, it's correct for 3...then 4, then 5...and so on, so this proves (if we can show the above equivalence), that this formula is correct for ALL n ≥ 1...actually we could show it's correct for (n ≥ 0...but NOT n < 0)
So plug in S(n) to the above and see if we can simplify to what the formula SHOULD give:
n(n + 1) / 2 + (n + 1)
--> factor out (n + 1)
(n + 1) * (n/2 + 1)
--> n/2 + 1 = n/2 + 2/2
(n + 1) * (n + 2) / 2
-->
This is indeed the formula for S(n + 1) given S(n) = n(n + 1) / 2.
Therefore this formula is correct for n ≥ 1 (again, we can use 0 as a base case and show it's correct for n ≥ 0...but if we use -1 as a base case, it does NOT hold and so this formula may not...and probably doesn't...hold for negatives)
n = 0:
sum should be 0:
0 * (0 + 1) / 2 = 0, CHECK
n = -1:
sum should be -1 + 0 = -1
-1 * (-1 + 0) / 2 = 0 ≠ -1, this is NOT a suitable base case...therefore this formula is NOT valid for n = -1 (and, I believe therefore NOT valid for ANY n < 0...negative numbers).
Edit:
There is one more somewhat direct way, to derive this:
You can assume that the function that describes the sum of the n is quadratic:
S(n) = an² + bn + c
Now, it's just a matter of finding the THREE coefficients: a, b, and c. Now you can reason that this WILL be the case (i.e. that it WILL be quadratic), because the differences between n and n + 1 is a constant...this means the function MUST be quadratic...btw, you can use this same technique to derive the formula for the natural numbers squared (by assuming a cubic)...then you could figure out the sum of the natural numbers cubed by assuming a quartic function, etc...