## Odd Sums of Consecutive Odds

Oscar Wilde’s character Algernon said in The Importance of Being Earnest, “One must be serious about something, if one is to have any amusement in life.” Of course in Wilde’s typical ironic fashion, Algernon was only referring to his own dedication to frivolous diversions. In that spirit, allow me a few moments to tell a story about one of the odder sums of odd integers I discovered as a kid.

I remember that sometimes when I was bored — most especially during long, bi-weekly car trips with my parents — I would play various games with integers. I have no idea why, but at one point I memorized some huge list of powers of 2 (I can still remember the list from 1 to 65,536). I also computed the squares, cubes, and so forth of most of the smaller integers. As a result, I discovered on my own quite a number of interesting patterns in the integers. I don’t remember most of them, but there is one in particular that has stuck with me through the years.

First, I can’t see how someone can become a mathematician and not know the relation

$\displaystyle{1 + 3 + 5 + ... + 2n-1= n^2}$

I discovered that on my own during one of those car trips. It was very exciting at the time, though I will admit the magic has dulled somewhat in the last 15 years. It was around then when I devoted some time to thinking about sums of odd integers. During some of that time, I noticed something peculiar hidden in the above sum.

$\displaystyle{1=1=1^3.}$

The sum of the next two odds is the second cube:

$\displaystyle{3+5=8=2^3.}$

The sum of the next three odds is the third cube:

$\displaystyle{7+9+11=27=3^3.}$

The sum of the next four odds is the fourth cube:

$\displaystyle{13+15+17+19=64=4^3.}$

The sum of the next five odds is the fifth cube:

$\displaystyle{21+23+25+27+29=125=5^3,}$

and so on.

That’s kind of surprising, right? Well, it’s less magical if you think about it. If this were true, we would immediately get the following equalities

$\displaystyle{1^3 + ... +n^3= \sum_{i=1}^{n(n+1)/2}(2i-1) = \left(\frac{n(n+1)}{2}\right)^2.}$

The second equality follows from our well-known formula for perfect squares. That the third term equals the first follows by a trivial induction argument. This first equality, though, gives us exactly our magic pattern for cubes since $n(n+1)/2 - n(n-1)/2 = n$. It is interesting to note that by the “Baby Gauss” formula, we have

$\displaystyle{1^3 + ... + n^3=(1+2+...+n) ^2.}$

It is easy to see that the pattern for general even powers of $n$ continues in a similar fashion since

$\displaystyle{\sum_{i=1}^{n^k}(2i-1)=n^{2k}.}$

The pattern for odds is similar in that you are adding strings of length $n^{(m-1)/2}$. However, you start skipping some integers. For $m=5$, we get

$\displaystyle{1=1=1^5.}$

Skip 3. The sum of the next four odds is

$\displaystyle{5+7+9+11=32=2^5.}$

Skip 13, 15, 17. The sum of the next nine odds is

$\displaystyle{19+21+23+25+27+29+31+33+35=243=3^5.}$

Skip 37, 39, 41, 43, 45, 47. The sum of the next 16 odds is

$\displaystyle{49+...+79=1024=4^5.}$

By slightly changing our point of view, we describe the pattern more concretely. Suppose the odd power in question is $m$. Consider the general sum of consecutive $n^{(m-1)/2}$ odd integers starting at $k_n:=k$ given by

$\displaystyle{k + (k+2) + (k+4) + ... +(k+2n^{(m-1)/2}-2).}$

We can shift $n^{(m-1)/2}-1$ from the last term to the first, $n^{(m-1)/2}-2$ from the next-to-last term to the second, and so forth, to re-write the sum as

$\displaystyle{(k+n^{(m-1)/2}-1) + ... + (k+n^{(m-1)/2}-1)}$

$\displaystyle{= n^{(m-1)/2}(k+n^{(m-1)/2}-1).}$

Since our sum should equal $n^m,$ this gives immediately that

$\displaystyle{k_n=n^{(m+1)/2}-n^{(m-1)/2}+1.}$

This formula agrees with what we have computed for $m=5$. Observe that if we hit every integer for each $m$, then we would get

$\displaystyle{k_n-k_{n-1}=2n^{(m-1)/2}.}$

which reduces to

$\displaystyle{(1+\frac{1}{n})^{(m-1)/2} = 1+\frac{1}{n},}$

and certainly only holds for $m=3$.

In the even case, one can see that regrouping the sums provides an iteration scheme to get the sum of the first $n$ of the $m^{\textrm{th}}$ powers in terms of the lower order sums. However, by what we’ve just described, the situation gets more complicated for odd powers, and so such an endeavor is probably better abandoned. Indeed, Jacob Bernoulli used an algebraic method to deduce a smarter way of computing them and produced a closed form in terms of Bernoulli numbers. That seems a much more reasonable approach. I found a quote attributed to him where he discusses his opinion of what I am guessing is the regrouping method I just described:

“With the help of [these formulas] it took me less than half of a quarter of an hour to find that the 10th powers of the first 1000 numbers being added together will yield the sum

91409924241424243424241924242500.

From this it will become clear how useless was the work of Ismael Bullialdus spent on the compilation of his voluminous Arithmetica Infinitorum in which he did nothing more than compute with immense labor the sums of the first six powers, which is only a part of what we have accomplished in the space of a single page.”

That’s got to sting a bit. And, mind you, Bernoulli and Bullialdus were contemporaries.

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### 19 Responses to “Odd Sums of Consecutive Odds”

1. D. Eppstein Says:

Wheatstone (1854) used the sum of consecutive odd numbers idea to prove the sum-of-cubes=squared-triangle formula. See, e.g., http://en.wikipedia.org/wiki/Squared_triangular_number

2. Gloria Rubi Says:

Esta bueno, léelo!

3. daniela Says:

muy interesante!!!!!

4. Sagaro Says:

odd sums and any function of adding sums of numbers is very interesting to me.
I would like to add something if i may be allowed.

I tend to look at numbers as building blocks to things and as they fit together, they seem to make more sense. the simple numbers 1 through 10 are not much use until they are assembled through equations and make something meaningful ( so my line of thinking goes).

The question began in my mind as to where the meaning begins? Are the first numbers as simple as I thought?

Lets take a simple example and change to something very impossibe to do.

I first began with the number 99 and decided to work backwards.
1/99 is shown to be a loop of 01 repeating itself. big deal. but i saw its close to 100, and 100 is 10 squared 10E2.
seeing that 2, made me think, here were the pieces of my puzzle.
100 – 1 is 99, 10 E2 is 100. If i were to start with 01, and multiply it by 01, keeping all digits ( including the zeroes before a number) and add the series i would get a hand written rough work that looked like
01
+ 0001
+ 000001
+ 00000001
____________
= 0101010101…….
ok there ya go.

98 is 100-2. 1/98 is start at 01 multiply by 02, one part that you may not have seen in the last example is the next sum is a multiplcative of the line above it, to be more clear…

01
+0002
+000004
+00000008
+0000000016
+000000000032
+00000000000064
+0000000000000128 <– the numbers begin to affect previous lines)
+000000000000000256
___________________
=010204081632653056 the total so far

with more lines you can get the exact value of 1/98.

two solid examples led me to this equation.

10E(C) – B/A = #

ok # sign is a number your looking for, previous examples were #= 98,99
C is the range,
(you see previously the numbers moved two spaces to the right?)
( if numbers only moved one space to rigth then C is 1)
( 3 spaces and C = 3, simply, the amount you wish to move right)
B is the multiplicative, how much you will multiply each number by
( 99 used B = 1, 98 used B = 2)
A is the number you want to start with, we used 1 in those examples, but it can change to any value you want to begin with.
TEST
we wish to begin with a number 14, multiply it by 2 every successive time, move 2 spaces to the right each time, what numbers inverse will that equate to?

[(10E2)-2] / 14 = 98/14 = 7… hence if we wrote it out the answer would be 1/7.
14
0028
000056
00000112
0000000224
000000000448
00000000000896
0000000000001792
000000000000003584
__________________
142857142857142784
two actual loops, and if you kept going it woudl repeat forever.
Has anyone known of this before I would really like to know.

Equation VALID

the impossible.
what if we multiply a number so greatly, that the left side of the stairs goes to the left and not a vertical line as we have seen until now.
start at 2 multiply by 20 and move one space to the right.

2
40
800
16000
320000
___________________________
if we kept this going, you can see right away, that you will never have a starting number,
but you will aquire a portion of the pattern some may say, NOT TRUE since the numbers keep stepping on other values no final portion of the number will be made. Ok so a real math genius would say, ya but the sum of 2 multiplied by 20 could be solved using some weird formula ( insert weird formula name here). 🙂
I would like to see that answer.

My ,method, C = 1 A = 2 B = 20
[(10E1) – 20]/ 2 = (10 – 20)/ 2 = (-5) what the heck does that mean???
1/5 is .2, is it not, by a little rearranging of the data in my mind, i cant say for real if sum of my little series is .2, but I simply stop and say, wow, that equation of mine seems to make negative numbers have a whole new meaning. Now where did I put that J operator???? is this just 2 E1/2? Just what does this mean, i need help, someone please respond.

I wish i had more time to discuss, but i dont want a long message and people just skim over it.
Thanks all,

Sagaro.
i made an msn space, feel free to check it out, I need some math friends

5. Peter Luthy Says:

Hi Sagaro,

Thanks very much for your post and interest in mathematics. Indeed you have found an interesting idea, one which we have discussed at great length here at the Everything Seminar. Everything you’ve proposed is a direct consequence of the following identity:

$\frac{1}{1-x}=1+x+x^2+x^3+x^4+...$

A non-rigorous proof of this fact comes from the following. If

$F=1+x+x^2+x^3+x^4+...,$

then

$xF=x+x^2+x^3+x^4+...=F-1.$

Solving for $F$ gives us

$F=1/1-x.$

This can be proven rigorously, so long as $-1, but the proof breaks down for any larger or smaller values of $x$. This is the key to your problem! Let’s check to see how this formula agrees exactly with what you’ve done!

To manipulate this to get your formula to get 1/99, notice that

$\frac{1}{99}=\frac{1}{100-1} =\frac{10^{-2}}{1-1/100}=10^{-2}(1 + 10^{-2} + 10^{-4} + ...)$

This equals

$10^{-2} + 10^{-4} + 10^{-6} + ...$

In your reasoning, $10^{-2} = 01$ and so we get that
01
+0001
+0000001
+…
—————
1/99

For $\frac{1}{7}$, observe that
$\frac{1}{7} = \frac{14}{100-2} = \frac{14\times 10^{-2}}{1-2/100}.$

Using our formula above, one can see that

$\frac{1}{7} = (14\times 10^{-2})(1+(2/100)+(2/100)^2+...$

which in your system of writing is just

14
+0028
+000056
+00000112
+…
—————
1/7.

So you see, your formula is correct for certain values of A,B, and C. If I’m understanding your final question here, I think this is the answer:

$\frac{1}{-5} = \frac{1}{(10-20)/2} = \frac{2}{10-20} = \frac{2\times 10^{-1}}{1-2}$.

We know that 2 is outside the allowed range for $x$. However, this formula still means something in terms of something called analytic continuation (where a function defined on a specific range of values is extended in a special way beyond where it was originally defined). In fact, one of the authors of this blog, Matt, has already written extensively about this subject in three posts on this blog, and he talks exactly about this new “meaning” behind this analytic continuation. Please confer with the first of his three posts about this at

https://cornellmath.wordpress.com/2007/07/28/sum-divergent-series-i/

6. Sagaro Says:

Peter I very much enjoyed your response to my question. thank you.. now part 2

What signifies the length of the repeating value for any inverse??
I have figured out the part of the non repeating part being zero for all prime and Prime X prime values..

Let me digress for a second if I may to explain what it is i am talking about for the newbies.

If you take a number and invert it, 7 inverted is 1/7 and has a value equal to, .142857142857142857…. forever. it has a loop of 6 numbers and no part that doesnt repeat. hence the term 0,6.
A number such as 1/12 is .083333333333. which conists of 2 non repeating numbers, and 1 repeating part. hence the term 2,1.

I have found out how the non repeating part grows but still remain stumped as to the repeating part.
Any ideas out there on this my friends?

If you care to ask about the growth of the non repeating part just ask, it is somewhat lengthy in explanation, but i love to share if needed

7. Peter Luthy Says:

Hi Sagaro,

Another interesting question. The question you’re asking is what a mathematician would say is “ill-posed.” For instance, we know that 1/9 is 0.1111, but is that 1 repeating or is it 11 or 111 or 1111 repeating? I can find a formula for “a length” of the repetition for the inverse of a number, but I’m not exactly sure how to reduce it to give you the shortest possible way in some nice, easy way.

Here is my idea. Suppose that you have a number $n$ which is not divisible by $2$ or $5$. Suppose that the decimal expansion of $n$ has a cyclic part of length $\beta$ and a non-cyclic part of length $\alpha$. Then if you think about it for a couple seconds, you’ll see that

$\frac{10^{\alpha +\beta}}{n} - \frac{10^{\alpha}}{n}$ is an integer.

In other words, we have that

$10^{\alpha + \beta} \textrm{ mod } n = 10^{\alpha} \textrm{ mod } n.$

Here the $\textrm{ mod }$ means modulo (from modular arithmetic).

Since $n$ is not divisible by 2 or 5, we can cancel the $\alpha$ out and deduce that

$10^{\beta} \textrm{ mod } n = 1 \textrm{ mod } n.$

In fact these steps all work backward in the sense that if $10^{x} \textrm{ mod } n = 1 \textrm{ mod } n,$ and $n$ has repeating length $\alpha$ then $\frac{10^{x+\alpha}-10^{x}}{n}$ must be an integer and so the decimal cycle of $n$ must be a divisor of $x$.

So how is this helpful? Well, all we need to do is compute the smallest positive power we need to raise 10 to to get 1 plus a multiple of n. This, of course, will give us the shortest possible length for the cycle. However, computing this particular power can be kind of hard. There are a couple of theorems which make life easier. For instance, there is a nice result called Fermat’s Little Theorem that says if $p$ is a prime then for any integer $a$, we have the relation $a^{p-1}=1 \textrm{ mod }p.$ Thus every prime must have a decimal cycle of length $p-1$. However, this is not a sharp result: for example, the 1/37 has a cycle of length 3 (which is indeed a divisor of 37-1=36).

The Swiss mathematician Euler later generalized this theorem. The generalization says that if $a$ and $n$ are coprime (meaning they share no common factors) then $a^{\phi(n)}=1 \textrm{ mod }n,$ where $\phi(n)$ denotes how many positive integers less than $n$ are coprime to $n$. However, this also is not a sharp result: $\phi (27)$ is quite a large number, but 1/27 has a cycle of length 3.

Having thought about this a little bit, I’m not sure how exactly to proceed beyond this point to a stronger theorem. But maybe since I’ve given you a little bit of information, you can figure it out and let me know!

–Peter

8. disquisitionesmathematicae Says:

This might help.
$a^{lcm[\phi(m_{1}),\phi(m_{2}),...,\phi(m_{n})]}\equiv 1\ (mod\ lcm[m_{1},m_{2},...,m_{n}])$

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