In my previous post you may have noticed that I neglected to prove one of the important results. One of my readers requested I post a formal proof. So here it is. Be warned, this is some serious stuff.

Consider the sequence of odd numbers:

1, 3, 5, 7, …

We want to prove:

Let’s start with something easier. For any arithmetic sequence:

a1, a2, a3, …, an

The sum of the first n terms is:

Instead of giving a proof for this lemma, I will elect to show a conceiving example.

How would we add up the numbers 1 to 100?

Write the sum:

S = 1 + 2 + 3 + … + 98 + 99 + 100

Write the sum in reverse:

S = 100 + 99 + 98 + … + 3 + 2 + 1

Add these two sums together:

2S = 101 + 101 + 101 + … + 101 + 101 + 101

Simplify:

2S = 100 * 101

Solve for S:

S = 100 * 101 ÷ 2

So in general:

Now using this formula we can prove the general result.

Hence:

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