# Odd Sums Revisited

In a previous post, I investigated the sequence of odd numbers:

$1, 3, 5, 7, 9, 11, \ldots$

In particular, I investigated the ratio between the sum of the first “k terms” and the sum of the next “k terms.” I found that the ratio was always 1/3. I attempted this investigation with the sequence of even numbers and the ratio failed to be constant.

I recently had a brain wave to continue the investigation. Here was my question: “what sequences of numbers have consistent ratios when you sum the first k terms and the next k terms?”

After a bit of investigation, I found another sequence!

$2, 6, 10, 14, 18, 22, \ldots$

Here is the ratio of the first term and the second term:

$\dfrac{2}{6} = \dfrac{1}{3}$

The ratio between the first 2 terms and the next 2 terms:

$\dfrac{2 + 6 }{10 + 14} = \dfrac{8}{24} = \dfrac{1}{3}$

The ratio between the first 3 terms and the next 3 terms:

$\dfrac{2 + 6 + 10}{14 + 18 + 22} = \dfrac{18}{54} = \dfrac{1}{3}$

We have a constant ratio! What is even more interesting is that this ratio is 1/3, just like before! Below is another sequence that works:

$3, 9, 15, 21, 27, 33, \ldots$

And another:

$4, 12, 20, 28, 36, 44, \ldots$

Here is one that uses decimals:

$1.1, 3.3, 5.5, 7.7, 9.9, 12.1, 14.3, \ldots$

$-5, -15, -25, -35, -45, -55, -65, -75, \ldots$

In fact, there are an infinite number of different sequences. All of these sequences share the exact same consistency as the original sequence. The ratio will always be 1/3.1 Insights like this are what make mathematics truly beautiful.1

1Below is the method I used to generate these new sequences. First, we need to represent the sequence in general. We will restrict ourselves to arithmetic sequences for this post. Hence, the general sequence can be written as:

$a, a+d, a+2d, a+3d, a+4d, \ldots$

We want our arithmetic sequence to have a constant ratio. We could rephrase this requirement as “the ratio of the first and second term must be equal to the ratio of the first 2 terms and the next 2 terms.” In symbols, this means:

$\dfrac{a}{a + d} = \dfrac{a + (a + d)}{(a + 2d) + (a + 3d)}$

Simplifying:

$\dfrac{a}{a + d} = \dfrac{2a + d}{2a + 5d}$
$a(2a + 5d) = (a + d)(2a + d)$

$2a^2+ 5ad = 2a^2 + 2ad + ad + d^2$

$2a^2 + 5ad = 2a^2 + 3ad + d^2$

$5ad = 3ad + d^2$

$0 = -2ad + d^2$

$0 = d(d - 2a)$

Since we want our sequence to be interesting, d ≠ 0 and a ≠ 0. Thus, d = 2a. Hence, any sequence that satisfies the condition d = 2a will exhibit our desired property.2

Now our general sequence looks like this:

$a, a+2a, a+4a, a+6a, a+8a, \ldots$

$a, 3a, 5a, 7a, 9a, \ldots$

From this, we see that the ratio will be:

$\dfrac{a}{3a} = \dfrac{1}{3}$

Since the a’s cancel, our general sequence reduces to our original sequence of odd numbers. Thus, all the properties that we discovered about the sequence of odd numbers will apply to all these new sequences. This also means that the only ratio ever possible is 1/3. Amazing!

2 I have another footnote for two reasons. First, I think it is pretty meta to put a footnote inside a footnote. Second, I actually only showed that the condition d = 2a will force the ratio of the first and second terms equal to the ratio of the sum of the first 2 terms and the next 2 terms.

I neglected to prove that this generalized to the sum of the first k terms and the sum of the next k terms. In other words, I didn’t prove that:

$\dfrac{a + (a + d) + (a + 2d) + \ldots + (a + (k - 1)d)}{(a + kd) + (a + (k + 1)d) + \ldots + (a + 2kd)} = \dfrac{1}{3}$

It should be obvious to the reader that if you replace d with 2a and factor out an a, the above ratio will reduce to:

$\dfrac{1 + 3 + 5 + \dots + (2k - 1)}{(2k + 1) + (2k + 3) + \dots + (4k + 1)}$

This is equivalent to the original ratio investigated in the first post.

If anyone is still reading at this point, they may notice that in the original post I didn’t prove that the above ratio was equal to 1/3. You would be correct. The proof is left as an exercise for the reader 😉