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Gottfried Helms science forum Guru
Joined: 24 Mar 2005
Posts: 301

Posted: Thu Jul 13, 2006 2:40 am Post subject:
Re: Odd Squares, 8 (2^3), and the Sum of the First n Positive Integers



Am 13.07.2006 03:58 schrieb rer:
Quote:  I did see it noted that the odd squares are 1 (mod
(http://mathworld.wolfram.com/SquareNumber.html). I imagine it is noted
elsewhere, but they also have a relationship to the sum of the first n
positive integers.
If it is noted that the sum of the first n positive integers =
n(n+1)/2, then
((2^3)(((n(n+1))/2))+1=(2n+1)^2
(8(n^2+n)/2)+1=4(n^2)+2n+2n+1
(4(n^2+n))+1=4(n^2)+4n+1
4(n^2)+4n+1=4(n^2)+4n+1

.......... = (2n +1)^2 by binomial expansion.
Gottfried Helms 

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rer science forum beginner
Joined: 10 Jul 2005
Posts: 21

Posted: Thu Jul 13, 2006 1:58 am Post subject:
Odd Squares, 8 (2^3), and the Sum of the First n Positive Integers



I did see it noted that the odd squares are 1 (mod
(http://mathworld.wolfram.com/SquareNumber.html). I imagine it is noted
elsewhere, but they also have a relationship to the sum of the first n
positive integers.
If it is noted that the sum of the first n positive integers =
n(n+1)/2, then
((2^3)(((n(n+1))/2))+1=(2n+1)^2
(8(n^2+n)/2)+1=4(n^2)+2n+2n+1
(4(n^2+n))+1=4(n^2)+4n+1
4(n^2)+4n+1=4(n^2)+4n+1
for example:
the sum of the first positive integer, 1, is 1
then, (8*1)+1=9 or ((2*1)+1)^2=3^2
the sum of the first two positive integers, 1+2, is 3
then, (8*3)+1=25 or ((2*2)+1)^2=5^2
the sum of the first three positive integers, 1+2+3, is 6
then, (8*6)+1=49 or ((2*3)+1)^2=7^2
the sum of the first four positive integers, 1+2+3+4, is 10
then, (8*10)+1=81 or ((2*4)+1)^2=9^2
the sum of the first five positive integers, 1+2+3+4+5, is 15
then, (8*15)+1=121 or ((2*5)+1)^2=11^2
etc. 

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