Math: The quadratic saga - Ep. 3

In the next day I dedicated some time to simplify the formula. I wanted to find something shocking, something that I would never imagine - and for that I had to make that formula smaller. So, we had:

n² = [2(n-1)+1] + [2(n-2)+1] + [2(n-3)+1] + ... + [2(n-n)+1]

We can see that in each "part" (part of the n-1, part of the n-2, etc) there is a +1 being added. So, for example, if we want to achieve the "part of (n-n)" starting from the "part of (n-1)", we are going to have a total of  'n'  parts. Let me give a example with numbers so that it is easier to understand. Let n=4:

4² = [2(3)+1] + [2(2)+1] + [2(1)+1] + [2(0)+1]

We can see that the number of "parts" is equal to 'n'. If we let n=4, for example, we are having 4 "parts", therefore since there is a '+1' in each part, in total we are having '+4'. So:

n = 4 || 4 "parts" || +1 for each part = total +4

And we can do that for any value of 'n':

n = 1 || 1 "part" || +1 for each part = total +1

n = 2 || 2 "parts" || +1 for each part = total +2

So we can relate:

n = n || n "parts" || +1 for each part = +n

So we can separate this '+1' of each part into one single letter: +n. Now we have:

n² = [2(n-1) + 2(n-2) + 2(n-3) + ... + 2(n-n)] + n

Now an easy simplification, separation of the '2' multiplying each "part":

n² = 2[(n-1) + (n-2) + (n-3) + ... (n-n)] + n

I am going to stop typing now. It isn't the same reason from the other episodes, this time I didn't stop until I reach my objective. The only problem was that... No, I am telling in the next episode. I will make the next episode different - I will divide it in two parts: the huge path that I made; and the simple answer that was right in front of me and I didn't see. The next episode is going to be the last episode of this "adventure"...