What is the name of this formula for finding the sum of consecutive numbers?

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The discussion centers on a formula for finding the sum of consecutive numbers, specifically illustrated through the difference of squares, where (x+1)^2 - x^2 equals 2x + 1. Participants note that while this formula is not widely recognized by a specific name, it relates to the well-known difference of two squares. The conversation also touches on the historical connection to Pythagorean discoveries, particularly the sum of odd numbers equating to a square. The formula demonstrates that the difference between the squares of consecutive integers yields the sum of those integers. Overall, the thread highlights the mathematical relationships and historical significance of these formulas.
RobinSky
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I found out that if you take for example 13^2-12^2 you get 13+12 and this works for all numbers, the formula is (x+1)^2-x^2=2x+1

I'm not sure if this formula is famous/known but I'm sure it is and my question is; is there a name for it and if so, what's the name for it?
 
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The more general form a2 - b2 = (a + b)(a - b) is just called 'difference of two squares' and is fairly constantly useful as is the more general difference of two n powers which gives you the formula of the sum of a geometric series. In your quoted cases the difference happens to be 1, but I don't know that there is a special name for that (I do know one is not needed).
 
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12^2 = 144
13^2 = 169

169 - 144 = 12 + 13 = 25

Then it just keeps on going, with every single numbers followed by each other.

100^2 = 10000
101^2 = 10201

10201 - 10000 = 100 + 101 = 201

Sure you knew that this was what I am talking about?:smile:
 
Maybe this helps as well, not sure but also a formula I came up with.

x^2+x+n=n^2

Let's now say x is 12 again. n = x+1

12^2+12+13=13^2

144+24+1=169.

What I'm trying to ask is, what's the name of this "thingy"(formula/teorem what ever)? If you choose any number for the variabel x, and then n=(x+1). Then the difference between the left & the right side is (x+n).

(n^2)-(x^2)=(x+n) there you go!
 
Hi RobinSky! Welcome to PF! :smile:

I'm afraid epenguin already gave the answer, but apparently you haven't matched his formula with your own yet.

Cheers! :smile:
 
Ahh yes, now I see it! Thanks anyway, fun though to see that I came up with this old formula in my own mind :biggrin:
 
It is simply (n+1)^2-n^2= n^2+ 2n+ 1- n^2= 2n+1

Or, as epenguin did it, (n+1)^2- n^2= (n+1+n)(n+1-n)= (2n+1)(1)= 2n+1.
 

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