# Math Theorem?

Who’s theorem (if there is one), or what formula satisfies:

2 squared (4) = 1 + 2 + 1

(don’t know how to write this algebraically)

ie. The sum of the numbers 1 through N + the sum of numbers 1 through (N-1) = N squared.

Actually, I’m pretty sure it’s a variant of the “arithmetic series” sum formula. See:

http://mathworld.wolfram.com/ArithmeticSeries.html

You can obtain your statement from the sum formula, by substracting n and shifting terms around. I can try to write it down if you wish, not sure how legibly it will come out without math symbols.

lemme see if this works for other values. I’m a math geek, but I’ve never heard of this

3

1+2+3+1+2=9

hmmm

4

1+2+3+4+1+2+3=16

hmm

10

1+2+3+4+5+6+7+8+9+10+9+8+7+6+5+4+3+2+1=100

It seems to check out.

To answer your question about algebraic notation, I would guess that it would be something like this:

n^2=(n[n-(n-1)]+n[n-(n-2)]…n[n-1]+n[n])+(=(n[n-(n-1)]+n[n-(n-2)]…n[n-2]+n[n-1])

Please bear in mind that the brackets are not part of the equation. I just donj’t know how to make subscripts, so that was the best that I could do. Anyone else’s help in symplifying the notation WOULD be appreciated.

Me bad. You asked who’s, well it would seem Gauss figured it out acording to my link (though I wouldn’t be surprised if there was someone him before who also did).

Trying to write it out then:

Sum (1,n) i + Sum(1,n-1) i = n^2
2 Sum (1,n-1) + n = n^2
Sum (1,n-1) = (n^2-n)/2

Now add n to both sides and simplify, and you get the sum formula. (Read “big Sigma” for “Sum” and the parameters go below and over the sigma letter, in notation.)

Geometrical proof:

Visualize the sum of 1 through N as a collection of dots in the shape of a right isosceles triangle. Visualize the sum of 1 through N-1 similarly. Join the two triangles along their hypotenuses, and the resulting figure is a square with N dots on each side.

Me bad. You asked who’s, well it would seem Gauss figured it out acording to my link (though I wouldn’t be surprised if there was someone him before who also did).

Trying to write it out then:

Sum (i=1,n) i + Sum(1,n-1) i = n^2
2 Sum (i=1,n-1) i + n = n^2
Sum (i=1,n-1) i = (n^2-n)/2

Now add n to both sides and simplify, and you get the sum formula. (Read “big Sigma” for “Sum” and the parameters go below and over the sigma letter, in notation.)

Me bad. You asked who’s, well it would seem Gauss figured it out acording to my link (though I wouldn’t be surprised if there was someone him before who also did).

Trying to write it out then:

Sum (i=1,n) i + Sum(1,n-1) i = n^2
2 Sum (i=1,n-1) i + n = n^2
Sum (i=1,n-1) i = (n^2-n)/2

Now add n to both sides and simplify, and you get the sum formula. (Read “big Sigma” for “Sum” and the parameters go below and over the sigma letter, in notation.)

Me bad. You asked who’s, well it would seem Gauss figured it out acording to my link (though I wouldn’t be surprised if there was someone him before who also did).

Trying to write it out then:

Sum (i=1,n) i + Sum(1,n-1) i = n^2
2 Sum (i=1,n-1) i + n = n^2
Sum (i=1,n-1) i = (n^2-n)/2

Now add n to both sides and simplify, and you get the sum formula. (Read “big Sigma” for “Sum” and the parameters go below and over the sigma letter, in notation.)

Hope at least the explanation was worth it, heh.

Subscript and Superscript, which should be included in the FONT or SIZE dropdowns, ahem, are accomplished as follows:

(sub)…(/sub)

(sup)…(/sup)

Just use brackets instead of parentheses.

Jestocost’s link discusses the formula for arithmetic sums in general. If you start with a and add terms that increase by d, until you have n terms (the last term is A=a+d(n-1) ), the sum total will be n/2(2a + d(n-1)).

In rpkc4’s case, 1+2+3…n, a=1, d=1, so the formula is just n/2(n+1). The formula for 1+2+3…n-1 is n/2(n-1). When you add those two together, you get n^2.

PS: A quick and dirty way of averaging an arithmetic series is to take the average of the first and last term, and multiply that by the number of terms. n(a+A)/2, in other words. Same thing.

Who’s [sic] theorem? It must be ancient. I am sure it is older than Gauss and not every theorem has a name. Although Pythagoras’s theorem has a name, the equally ancient facts that there are infinitely many primes and that the square root of 2 is irrational doesn’t.

The there’s Chebychev’s theorem:

Chebychev proved it,
You could too.
There’s always a prime
'Twixt n and n times 2.

G’day

Arrange n-squared objects into a square. Look at the rows that are parallel with a main diagonal. The first on is a corner, and contains one object, the next contains two, the next three… The nth is the main diagonal, and contains n. Then they count down to the opposite corner, with one again.

Is anybody surprised that the nth triangular number plus the (n-1)st withthe terms written in reverse order gives the nth square?

Regards,
Agback