This month I am going to be discussing some of the more interesting things I learn in my math classes (I am taking two) and would like to begin with a seemingly simple question…
What is the sum of the first n integers?
Well, uhm. How would you solve this? We could easily go out and create a list of the numbers we want to count, say to 5…
1 + 2 + 3 + 4 + 5 = x
Well, we can do some nifty algebraic grouping and draw a conclusion that way as well…
(1 + 2) + (3 + 4) + 5 = 3 + 7 + 5 = 10 + 5 = 15
But that doesn’t actually answer the question. This is for a finite quantity, but what about finding for 500? What do you do now? You could sit for the next thirty minutes trying to organize the numbers in such a way that things are clean, but then again what about 5000? What about a million? Things get extremely complicated, or tedious, when you get out to that distance. In a moment we will be discussing how we can solve this problem inductively, but first ill share a bit of information about intuitively solving it.
Lets try to do some counting…
1 + 2 + 3 + … + n
We can see from this that there will be a pattern. Actually we can’t, but let me expand it a bit.
1 + 2 + 3 + 4 + … + (n3) + (n2) + (n1) + n
That should shed some light on the topic. If we look at the numbers we can pair them, as follows:
(n + 1) + [(n 1) + 2] + [(n – 2) + 3] + [(n – 3) + 4] + …
And you can see that these all simplify down quite well.
(n + 1) + (n + 1) + (n + 1) + (n + 1) + …
So counting n items is just a matter of pairing them up and adding the values together. Following that line of reasoning, we will find that there are half as many elements to count as are given. This train ends at the conclusion that …
[n(n + 1)]/2
Is the equation we are looking for. Lets give it a try for a number we already know…
[5 (5 + 1)] /2 = (5 * 6)/2 = 30/2 = 15
Yes sir it seems to work.
So Inductive proof, how do those things work?
 Show it works for 1
 Assume it works for k
 Show that it works for k + 1
Not with the detail of course, but this is what it comes down to, according to my instructors this quarter =).
 [1 (1 + 1)]/2 = (1 * 2)/ 2 = 1
 [k (k+1)]/2

{(k+1)[(k+1)+1]}/2 = [k(k+1)]/2 + (k+1)
 [k2+k]/2 + (k + 1) – Distribution
 [k2+k+2k+2]/2 – Common Denominator
 [k2+3k+2]/2 – Simplification
 [(k+1)[k+2)]/2 – Factoring
 {(k+1)[(k+1)+1]}/2 – Format Change
As you can see they are the same. I took the liberty of including the reason for the changes I was making, mainly because I don’t know how mathematically literate you all are. I will assume the best though =)