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 October 15th, 2012, 11:23 AM #1 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 sum of integers squared i have been searching for a formula for this but i could only find one for the sum of the first n integers squared which was $\frac {n(n+1)(2n+1)}{6}$ now i have been producing a formula that can calculate the sum of integers squared no matter where you start or the difference between the integers. i became: $=\frac {n(t_n+t_1)}{2}\quad -\quad \left(sqrt (t_n)-sqrt (t_1)\right)^2\quad * \quad\frac{n(n-2)}{6n-6}$ so for example if you want to calculate $9+25+49+81+121=285$ you have to do: $=\frac {5(121+9)}{2}\quad -\quad \left(11-3\right)^2\quad * \quad\frac{5(5-2)}{6*5-6}$ $=\frac {5*130}{2}\quad -\quad64\quad * \quad\frac{5*3}{24}$ $=325\quad -\quad\frac{15*64}{24}$ $=325\quad -\quad40$ $=285$ is this anything special or anything new that might be usefull. please comment.
 October 15th, 2012, 12:32 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: sum of integers squared If the sum of the squares of the first n positive integers is f(n), then the sum of the squares of k+1, k+2, ..., n is f(n) - f(k). Concretely: (2n^3 + 3n^2 + n - 2k^3 - 3k^2 - k)/6
October 15th, 2012, 02:15 PM   #3
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Joined: Dec 2006
From: Lexington, MA

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Re: sum of integers squared

Hello, gelatine1!

Quote:
 I have been searching for a formula for this, but i could only find one for the sum of the first n integers squared which was: $\frac {n(n+1)(2n+1)}{6}$ Now i have been producing a formula that can calculate the sum of integers squared, no matter where you start or the difference between the integers.

With a constant difference between the integers, we have an arithmetic sequence.

$\text{Then: }\:S \;=\;a^2\,+\,(a+d)^2\,+\,(a+2d)^2\,+\,(a+3d)^2\,+\ ,(a+4d)^2 \,+\,\cdots\:+\,(a+nd)^2$

$\;\;\;\text{where }a\text{ is the first term, }d\text{ is the common difference,}
\;\;\;\text{and there are }n+1\text{ terms.}$

$\text{Add these squares:}$

[color=beige]. . . [/color]$\begin{array}{c} a^2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \\
\vdots \;\;\;\;\;\;\; \vdots \;\;\;\;\;\;\; \vdots \\

$\text{And we have: }\:S \;=\;(n+1)a^2\,+\,n(n+1)ad\,+\,\frac{n(n+1)(2n+1)} {6}d^2$

[color=beige]. . . . [/color]$\text{ta-}DAA!$

October 15th, 2012, 07:23 PM   #4
Math Team

Joined: Jul 2011
From: North America, 42nd parallel

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Re: sum of integers squared

Quote:
 Originally Posted by gelatine1 i have been searching for a formula for this but i could only find one for the sum of the first n integers squared which was $\frac {n(n+1)(2n+1)}{6}$ now i have been producing a formula that can calculate the sum of integers squared no matter where you start or the difference between the integers. i became: $=\frac {n(t_n+t_1)}{2}\quad -\quad \left(sqrt (t_n)-sqrt (t_1)\right)^2\quad * \quad\frac{n(n-2)}{6n-6}$ so for example if you want to calculate $9+25+49+81+121=285$ you have to do: $=\frac {5(121+9)}{2}\quad -\quad \left(11-3\right)^2\quad * \quad\frac{5(5-2)}{6*5-6}$ $=\frac {5*130}{2}\quad -\quad64\quad * \quad\frac{5*3}{24}$ $=325\quad -\quad\frac{15*64}{24}$ $=325\quad -\quad40$ $=285$ is this anything special or anything new that might be usefull. please comment.
I think your formula is elegant, i've never seen it before, but i am not an expert and i have never had anything published of any consequence. I have published problems and solutions in School Science and Mathematics Association journal, editor, Ted Eisenberg.

http://www.ssma.org/publications

Read this link carefully and look at previous issues and problem proposal publications. If you can write up an article, maybe Mr. Eisenberg will publish it, although i suspect it is harder to publish articles, you may need credentials, but almost anyone can propose a problem.

Here is a possible approach, you send Ted Eisenberg your formula along with a numerical example, if you have a proof, send that too, you ask him to put it in the problem section of the journal, something like 'prove the following formula for the sum of n consecutive squares of integers with common difference starting at any integer square' or you may have a better title. Ted puts it in the journal, along with your name as proposer. Mathematicians all over the world see it and work on it, send in solutions. A few months after proposal, Ted publishes the best solution and mentions every name that sent in correct solutions.

I don't know how many mathematicians read this journal but it's got to be at least thousands of professionals worldwide so you will get some glory which you can use in a resume.

Like i said, i don't know how important your formula is, so i don't know if i'm giving you sound advice, you have to decide for yourself. In my opinion, Ted Eisenberg will put it in the journal.

 October 15th, 2012, 10:59 PM #5 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 Re: sum of integers squared i will give it a try. i just have to send a mail to $eisenbt@013.net$ for that ? or is there something on the site for that?
 October 16th, 2012, 12:22 PM #7 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 Re: sum of integers squared I've sent a huge email to him now I will let you know if he replies something.

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