Found that link!
Here it is:

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

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Regards,

RD
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Please keep all correspondence within the Group, so all may benefit !
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"Gord Dibben" <gorddibbATshawDOTca wrote in message
...
RD

I will ponder this for a while then quietly drift off to whatever planet I
currently reside on. Depends upon which set of meds I'm on<g

But I am improving.....honestly!

Thanks for the work you put into this explanation.

Saved for future regurgitation.


Gord

On Sat, 21 Oct 2006 12:56:31 -0700, "RagDyeR" wrote:

Well Gord, with no intention of implying that I'm well versed in
mathematics, I just happen to take a liking to this "Arithmetic Series".

A year or so ago, Dana put me on to a link (which I can't find at the
moment) that described how Gauss formulated this "truth".

In words it's:

<<<"The count of sequential numbers times the average of the first and last
number."

=A1*(A1+1)/2

So, the formula I posted is supposed to be "intuitive" for *ALL*
possibilities.

Here, the OP's request was for a sum of a series starting (or ending) with
1, therefore, the last (or first) number in this series is the *actual*
amount (count) of numbers in the sequential series, so your A1^2 works.

BUT ... what if the we're looking for, say 5 to 15?
A1 = 15
B1 = 5

In the original formula, A1 now has to become (A1-B1+1)
And the balance becomes (A1+B1)/2

So we now have:
=(A1-B1+1)*(A1+B1)/2

Taking it a step further, there's *no* stipulation that the sequential
series must have an increment or decrement constant of *1*.

Let's take the same 5 to 15, but here we want an increment constant of 5,
the SUM of the sequential series of 5, 10, &15.
A1 = 15
B1 = 5
C1 = 5

=((A1-B1)/C1+1)*(A1+B1)/2

So, that's why:
<<<"The count of sequential numbers times the average of the first and last
number."
=A1*(A1+1)/2

Is probably easier to relate to.

Gord Dibben MS Excel MVP