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Subject:  Gauss's formula
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Gauss's "theorem" says:

i=n
SIGMA i = n(n+1)/2
i=1

i.e., 1 + 2 + ... + n = n(n+1)/2

Someone in class asked whether this formula holds when i starts at i=0.
I said that it did, and I was correct, but...

I also said that it would work for i starting at i=2, and, although
there's a sense in which that's correct, there's a more important sense
in which it isn't correct...and neither is the claim about i=0.

To see what the formula really is, let's take a concrete case:

The sum of the integers from 1 to 6 can be found using Gauss's trick as
follows:

1+2+...+6
6+5+...+1
---------
7+7+...+7  

Then note that we have 6 7s, which is double the sum that we want,
so we get:

6*7/2 = 21 = 1+2+...+6

Let's try the sum of the integers from 0 to 6:

0+1+2+...+6
6+5+4+...+0
-----------
6+6+6+...+6

We have 7 6s, so we get 7*6/2 = 21.  

Note that it *looks* like the same formula as before.  But looks can be
deceiving.  Let's try the sum of the integers from 2 to 6:

2+3+...+6
6+5+...+2
---------
8+8+...+8

We have 5 8s, which is double the sum that we want, so we get:

5*8/2 = 20 = 2+...+6

Let's try the sum of the integers from 3 to 6:

3+4+5+6
6+5+4+3
-------
9+...+9

That's 4 9s, so 4*9/2 = 18 = 3+4+5+6.

Now let's look at the patterns we have:

integer range     formula
-------------     -------
0 to 6            7*6/2
1 to 6            6*7/2
2 to 6            5*8/2 (do you begin to see a pattern?)
3 to 6            4*9/2

I predict that the sum of the integers from 4 to 6 will be 3*10/2.
(Try it.)

How can we write that formula more generally?
We know that Gauss's formula for the sum from 1 to 6 is 6(6+1)/2.
In general, the sum from 1 to n (for any n) is n(n+1)/2.

Note that the sum from 2 to 6 is (6-1)*(6+2)/2.
And the sum from 3 to 6 is (6-2)*(6+3)/2.
We can now see that the sum from 1 to 6 is really (6-0)*(6+1)/2.
And the sum from 0 to 6 is really (6-(-1))*(6+0)/2.

We can generalize:  The sum of the integers from j to n (for any j,n)
i.e., j + (j+1) + (j+2) + ... + n; i.e.:

i=n
SIGMA i
i=j

is (n-(j-1))*(n+j)/2.

So, in particular,

i=n
SIGMA i = (n-(2-1))*(n+2)/2 = (n-1)(n+2)/2
i=2

Challenge:  Use mathematical induction to prove that formula for 2+...+n.

(Sounds to me like a wonderful final-exam problem :-)