# Odd Sums Re-Revisited

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:

$\displaystyle\sum_{i=1}^{k} a_i \div \displaystyle\sum_{i=k+1}^{2k} a_i = \dfrac{1}{3}$

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

a1, a2, a3, …, an

The sum of the first n terms is:

$S_n = \dfrac{n(a_1+a_n)}{2}$

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:

$S_n = \dfrac{n(a_1+a_n)}{2}$

Now using this formula we can prove the general result.

$\displaystyle\sum_{i=1}^{k} a_i = \dfrac{k(1+(2k-1))}{2} = \dfrac{k(2k)}{2} = k^2$

$\displaystyle\sum_{i=k+1}^{2k} a_i = \dfrac{k((2k+1) +(4k-1))}{2} = \dfrac{k(6k)}{2} = 3k^2$

Hence:

$\displaystyle\sum_{i=1}^{k} a_i \div \displaystyle\sum_{i=k+1}^{2k} a_i = \dfrac{k^2}{3k^2} = \dfrac{1}{3}$

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