How do I derive the formula for sum of n consecutive natural numbers

Well the only "direct" way I know of would require writing down the recursion relation, then, solving it...which will likely not make sense to you (unless you are familiar with differential equations or are at least past Cal. 3 in college).

Now there are ways to reason about it, and then you CAN mathematically prove that your reasoning is correct:

If we sum the 1st 100 numbers...then we will notice the following pattern:

1 + 100 = 101
2 + 99 = 101
3 + 98 = 101
4 + 97 = 101
...

So, for 1 through 100, how many of theses 101's do we add?

Well, if you think about it it's easy to figure out (although I DID have to think for a second):

Let's guess that there are 50 of these terms:

so what plus 50 is 101?

50 + x = 101 --> x = 101 - 50 = 51
-->

So for the first 50 numbers, the 50 (50), will be paired with 51...there are FIFTY 101's.

Ok, this isn't really all that convincing yet...but you DO notice that 50 = 100 / 2, right? And, in fact, it stands to reason that since we split the 100 numbers into TWO, that there would be a total of HALF (100/2 = 50) sums (of the 101)...

But that's all well and good for even numbers...what about odds?

Well let's try a much simple addition: 1-5:

Here we have:

1 + 5 = 6
2 + 4 = 6
3 + 0 = 3
-->

TWO sums of 6 and ONE of 3...totaling 15. Now, how do we handle that? Does the above work (if you take half the number, then multiply by one more?):

take half of 5 = 2.5....NOT an integer...weird, huh?

Well let's tread on, multiply by 6 (5 + 1):

6 * 2.5 = 6 * 5/2 = 3*5 = 15, but that's the RIGHT answer!!!

maybe we are on to something...maybe the formula is:

S = n * (n + 1) / 2
-->

Let's check by mathematical induction (this is a formal proof that the above formula IS correct):

Let's FIRST check the base case: