Study Curve

The below answer is very nice for this problem in particular, but here is a more general approach that sounds like what you were aiming for:

First, show that this is true for n = 1, i.e. 1 = (1∙2)/2. In this case you "show" this just by noting that the equality holds, but sometimes there is a little more work to be done.

Then, assume that it is true for n = k, and show that it is true for n = k + 1. For this problem in particular, you would assume that 1 + 2 + 3 + ⋯ + k = (k∙(k + 1))/2, and then show that 1 + 2 + 3 + ⋯ + k + 1 = ([k + 1]∙([k + 1] + 1))/2.

Once you've shown these two things, you can conclude by the principle of induction that this formula is true for all n. Intuitively, you can think of this via the ladder analogy: if you can reach the first step of the ladder, and from any given step, you can reach the next step, then you can reach any step on the whole ladder. In terms of n, you show that the formula holds for n = 1, so then because 2 = 1 + 1 it holds for n = 2, and because 3 = 2 + 1 it holds for n = 3, and so on…

I think the way you can prove this one in particular is to add the number together forward and backwards.

1 + 2 + 3 .... + n
+
n + (n - 1) + (n - 2) ... + 1

= (n + 1) + ((n-1) + 2) + ((n - 2) + 3 ... (n + 1)
= (n + 1) + (n + 1) + (n + 1) ... (n + 1)
= (n + 1) * n

Since each number was added twice

n (n + 1) / 2